UNIVERSITY  OF  CALIFORNIA 
AT    LOS  ANGELES 


s 


THE    CALCULATIONS 


OF 


ANALYTICAL  CHEMISTRY 


With  Compliments  of 


The  Author. 


9eb  fiorh. 
THE  MACMILLAN  COMPANY 

LONDON :  MACMILLAN  &  CO.,  LTD. 
1900 


All  rights  reserved 


THE    CALCULATIONS 


OF 


ANALYTICAL  CHEMISTRY 


EDMUND  H.  MILLER,  PH.  D., 

INSTKUCTOB     IN     ANATA'TICAL     CHEMISTRY    AND     ASSAYING 
COLUMBIA    UNIVERSITY 


THE  MACMILLAN  COMPANY 

LONDON  :  MACMTLLAN  &  CO.,  LTD. 

1900 

/' 

All  right*  reserved 


Entered  according  to  Act  of  Congress,  in  the  year  1900, 

BY  EDMUND    H.    MILLER, 
In  the  Office  of  the  Librarian  of  Congress,  at  Washington,  D.  C. 


PRKSS  OF 

GEORGE  G.  PECK 

NEW  YORK 


M(b  i 


PREFACE. 


THIS  text-book  is  intended  for  use  in  scientific  schools  and 
colleges,  in  connection  with  courses  in  analytical  chemistry,  and 
aims  to  give  a  logical  treatment  of  the  calculations  required  of 
an  analyst.  As  analytical  chemistry  is  so  closely  allied  with 
the  other  branches  of  chemistry,  many  of  the  calculations  given 
will  be  found  of  general  application.  The  matter  has  been 
selected  so  as  not  to  encroach  on  the  domain  of  physical  chem- 
istry but  is  intended  to  lead  up  to  the  study  of  that  subject. 

The  object  has  been  to  furnish  a  text-book,  which  shall 
give  the  necessary  information  concerning  those  important 
chemical  calculations,  which  every  student  should  thoroughly 
understand  before  taking  up  advanced  work.  The  method  of 
presentation  and  arrangement  differs  materially  from  the  old 
books  on  chemical  calculations.  Formulae  have  been  avoided, 
except  in  the  last  chapters,  so  that  the  student  shall  consider 
each  problem  individually  and  solve  it  from  a  knowledge  of 
chemical  laws  instead  of  substituting  in  formulae  for  different 
cases.  If  the  student  has  in  this  way  obtained  ability  to  apply 
the  laws,  he  can  readily  construct  a  formula  or  table  for  con- 
stant use. 

More  than  two  hundred  examples  are  given,  with  answers  ; 
all  of  which  can  be  solved  by  arithmetic  or  algebra.  They 
have  been  made  as  practical  as  possible  in  order  to  connect 
chemical  laws  with  practical  work.  Numerous  tables,  which 

»  210977 

270 


iv.  PREFACE. 

have  been  carefully  verified,  will  be  found  at  the  back  of  the 
book. 

The  assistance  of  Prof.  Freedman  of  the  University  of  Ver- 
mont on  Chapter  X.  and  of  Dr.  Joiiet  in  correcting  proof  and 
verifying  examples  is  gratefully  acknowledged.  Any  corrections 
or  suggestions  will  be  much  appreciated. 

E.   H.   M. 
HAVEMEYER  HALL, 

January,  1900. 


TABLE  OF  CONTENTS. 


CHAPTER  I. 

CALCULATION  OF  CHEMICAL  EQUIVALENTS  AND  ATOMIC  WEIGHTS. 

PAGE. 

Determination  of  Equivalents 

Two  Equivalents  of  Copper 4 

Electrolytic  Determination  of  Equivalents 4 

Definition  of  Atomic  Weight 5 

Relation  of  Chemical  Equivalents  to  Atomic  Weights 6 

Examples 9 

-      CHAPTER   II. 

CALCULATION  OF   FORMULAE  AND   PERCENTAGE. 

Empirical  Formula  from  percentage 12 

Formulae  of  Minerals 13 

Oxygen  Ratio 16 

Percentage  Composition  from  Formula 17 

Percentage  to  the  Dry  Basis 17 

Examples 19 

CHAPTER    III. 

CALCULATIONS  OF   MIXTURES   HAVING  A  COMMON  CONSTITUENT. 

/Chlorine  and  Bromine  in  Mixed  Silver  Salts 22 

Sodium  and  Potassium  in  Mixed  Sulphates 24 

Bicarbonate  and  Carbonate  in  Commercial  Bicarbonate 25 

Formation  of  Mixtures  of  Definite  Composition 27 

Examples 29 

CHAPTER   IV. 

CALCULATIONS  FROM   EQUATIONS. 

Quantitative  Meaning  of  Equations 32 

Weight  of  a  Constituent  in  a  Given  Weight  of  a  Compound 33 

Weight  of  a  Constituent  not  in  the  Substance  Weighed 35 

Quantity  of  Reagent  Necessary  to  Complete  a  Reaction 36 

Choice  of  Most  Economical  Reagent 88 

Examples 39 


vi.  TABLE   OF   CONTENTS. 

CHAPTER   V. 

v         CALCULATION  AND  USE  OF  FACTORS.  PAGE_ 

Calculation  of  Factors 42 

Use  of  Factors  in  Gravimetric  Analysis 42 

Use  of  Factors  in  Volumetric  Analysis 45 

Assay  Ton  System 46 

Examples 48- 

CHAPTER    VI. 

CALCULATIONS  OF  VOLUMETRIC  ANALYSIS. 

Discussion  of  Normal  Solutions 51 

Calculations  of  Alkalimetry  and  Acidimetry 57 

Permanganate  Calculations 60- 

^   Iodine  Calculations 64 

Adjustment  of  Solutions 66 

Subtraction  of  Excess  Necessary  to  Affect  an  Indicator 68 

Examples 70 

CHAPTER   VII. 

CALCULATIONS  OF  DENSITY  OK  SOLIDS  AND   LIQUIDS. 

Density  of  Solids 76 

Reduction  to  Density  at  4°  C 81 

Density  of  Liquids 81 

Dilution  to  a  Certain  Specific  Gravity 83 

Changes  in  Specific  Gravity  Due  to  Temperature 84 

Volume  Occupied  by  Precipitates 85 

Examples 88 

CHAPTER  VIII. 
CALCULATIONS  OF  GASES. 

Corrections  for  Temperature  and  Pressure 91 

Density  of  Gases 93 

Relation  between  Density  and  Molecular  Weight 94 

Density  by  Methods  of  Dumas,  Meyer  and  Hofmann 95 

Density  by  Effusion 97 

Correction  of  Weighings 98 

Calculations  of  Gas  Analysis 101 

Examples 105 


TABLE   OF   CONTENTS.  yii. 


CHAPTER    IX. 

CALCULATIONS  OF  CALORIFIC   POWER.  PAGE. 

Thermo-Chemical  Reactions 109 

Heats  of  Formation 110 

Units  of  Calorific  Power 112 

Calorific  Power  of  Solids  from  the  results  of  ultimate  analysis 114 

Calorific  Power  of  liquids 117 

Relations  of  Units  used  for  Gases 118 

Calorific  Power  of  Gases  from  the  results  of  analysis 119 

Calorific  Power  from  weight  of  lead  reduced 120 

Examples 122 

CHAPTER  X. 

ELECTRIC  AND  ELECTROLYTIC  CALCULATIONS  FOR  DIRECT  CURRENTS. 

Electrical  Units 124 

Calculation  of  Resistance 1 25 

Ohm's  Law 127 

Joule's  Law 128 

Mechanical  and  Electrical  Units  of  Power 130 

Electrolysis 131 

Kathions  and  Anions 131 

Faraday's  Laws 132 

Electro-chemical  Equivalents 133 

Formulae  for  Weight  of  Deposit,  Current  and  Power 133 

Counter-Electromotive  Force 135 

Thompson's  Law 137 

Ohm's  Law  for  Electrolysis 140 

Selective  Electrolysis 141 

Relations  between  Voltage  and  Current  in  Electrolytic  work 142 

Effect  of  Dissolving  Anodes 144 

Current  Density 145 

Examples,  With  Solution  of  Each 146 

TABLES. 

Weights 157 

Measures .• 158 

Atomic  Weights 159 

Factors 160 

Specific  Heat  of  Solid  Elements 162 

Conversion  of  Centigrade  and  Fahrenheit  Degrees 163 

Values  of  Normal  Solutions.  ..  ..  164 


viii.  TABLE   OF   CONTENTS. 

PAGE. 

Values  of  Tenth  Normal  Solutions 165 

Comparison  of  Degrees  Baume  with  Specific  Gravity 166 

Volume  of  One  Gram  of  Water  at  Different  Temperatures 167 

Density  of  Gases  and  Weights  of  One  Liter 168 

Vapor  Pressure  of  Water 169 

Specific  Gravity  and  Percentage  of  Ethyl  Alcohol 170 

Percentage  and  Specific  Gravity  of  Sulphuric  Acid 171 

Percentage  and  Specific  Gravity  of  Nitric  Acid 172 

Specific  Gravity  and  Percentage  of  Hydrochloric  Acid. . .    173 

Specific  Gravity  and  Percentage  of  Ammonia 174 

Relations  between  Units  of  Electricity,  Heat  and  Power 175 

Electro-Chemical  Equivalents 176 

Heats  of  Combination,  per  Gram-Equivalent 177 

Specific  Resistance  to  Electricity 178 

Calculated  and  Observed  Decomposition  Voltages 179 

Logarithms 180 


CHAPTER  I. 

'CALCULATION     OF     CHEMICAL     EQUIVALENTS     AND    ATOMIC 
WEIGHTS. 

CHEMICAL  EQUIVALENTS. 

THE  calculations  of  analytical  chemistry  are  based 
on  the  constant  relations  existing  between  the  different 
elements  in  a  chemical  compound.  These  relations  are 
best  expressed  by  equivalents,  that  is,  the  weight  of  the 
element  equivalent  to  one  part  of  hydrogen.  They  are 
the  equivalent  quantities  which  will  enter  into  reaction 
with  one  part  by  weight  of  hydrogen. 

For  example,  if  we  decompose  water  by  an  electric 
current,  we  have  hydrogen  and  oxygen  given  off  and  for 
every  gram  of  hydrogen  we  find  eight  grams  of  oxygen, 
therefore  the  equivalent  of  oxygen  is  eight,  for  eight 
parts  by  weight  of  oxygen  combine  with  one  part  by 
weight  of  hydrogen  to  give  nine  parts  of  water,, 

The  equivalent  of  copper  may  be  obtained  as  follows: 
When  pure  black  oxide  of  copper  is  heated  in  a  stream 
of  hydrogen,  the  oxide  is  reduced  to  metallic  copper  and 
the  oxygen  combines  with  the  hydrogen  to  form  water, 
which  can  be  absorbed  by  a  weighed  calcium  chloride 
tube  and  reweighed  after  the  experiment. 

Suppose  we  take  four  grams  of  black  oxide  of  copper 
and  carry  on  the  experiment  until  no  more  water  is 
formed:  then  weigh;  the  metallic  copper  remaining  is 


2  CHEMICAL  EQUIVALENTS. 

3. 196  grams :  the  loss  in  weight  is  the  oxygen  which  was 
combined  with  the  copper ;  so,  as  we  know  the  equiva- 
lent of  oxygen  to  be  8,  if  we  make  the  proportion  3.196 
(the  weight  of  copper)  10.804  (the  weightof  oxygen) : :  x:  8, 
x  will  be  the  equivalent  of  copper,  31.8.  Or  weigh  the 
water  formed  (the  increased  weight  of  the  calcium 
chloride  tube) :  this  is  0.9045  gram.  From  this  we  can 
obtain  a  direct  comparison  between  hydrogen  and  cop- 
per;  for,  of  the  0.9045  gram  of  water,  0.8040  gram  is 
the  oxygen  lost  by  the  copper  oxide,  so  0.9045-0.8040 
or  0.1005  gram  is  the  weight  of  hydrogen  which  has 
combined  with  the  oxygen  previously  in  combination 

with  copper;  hence3'19    or  31.8  is  the  weight  of  copper 

equivalent  to  one  gram  of  hydrogen  or  the  chemical 
equivalent  of  copper. 

To  obtain  the  equivalent  of  bismuth,  the  metal  can  be 
treated  with  nitric  acid  until  it  is  completely  converted 
to  oxide  and  after  ignition  weighed.*  3.6770  grams  of 
bismuth  gave  4.1016  grams  of  oxide.  The  increased 
weight  0.4246  gram  is  the  oxygen  combined  with 
3. 6770  grams  of  bismuth,  hence  the  equivalent  of  bismuth 
will  be  found  from  the  proportion 

3  6770  :  0.4246 :  :  x :  8.  x  =  69.27. 

If  we  heat  finely  divided  bismuth  with  an  excess  of 
sulphur  and  add  small  portions  of  sulphur  until  the  bis- 
muth is  completely  converted  to  sulphide,  then  remove 
the  excess  of  sulphur  by  digesting  with  a  caustic  alkali 
solution,  we  can  from  the  increased  weight  find  the 
equivalent  of  sulphur.  For  example,  3  grams  of  bismuth 
gave  3.6928  grams  of  sulphide.  To  get  the  equivalent 
of  sulphur.  3  :  0.6928  ::  69.27  :  x.  x  =  16. 

*  Schneider,  J.  prakt.  Chem.  [2],  50,461. 


CHEMICAL  EQUIVALENTS.  3 

We  have  now  the  following  equivalents:  Oxygen,  8; 
Copper,  31.8;  Bismuth,  69.27;  Sulphur,  16.  To  obtain 
that  of  silver,  heat  together  pure  finely  divided  silver 
and  an  excess  of  sulphur,  volatilize  the  excess  of  sulphur 
and  weigh  the  resulting  silver  sulphide.  Suppose  these 
to  be  the  results  :  4  grams  of  silver  gave  4.592  grams  of 
sulphide  ;  then  0.592  :  4  :  :  16  :  x  x  =  108,  the  equivalent 
of  silver.  Now  that  we  have  obtained  the  equivalent 
of  silver,  we  can  combine  it  with  a  number  of  elements 
and  get  their  equivalents.  For  example,  dissolve  three 
grams  of  silver  in  nitric  acid  and  pass  in  chlorine  until 
the  silver  is  completely  precipitated  or  pass  chlorine  over 
the  silver  heated  in  a  combustion  tube  until  it  is  com- 
pletely converted  into  chloride:  weigh  the  silver  chloride; 
weight,  3.9861  grams.  To  calculate  the  equivalent  of 
chlorine  we  have  0.9861  :  3  :  :  x  :  108.  x  =  35.5.  In  a 
similar  way  the  equivalents  of  bromine  and  iodine  can 
be  found. 

This  process  of  obtaining  one  equivalent  from 
another  can  be  carried  on  indefinitely  until  those  of  all 
the  known  elements  have  been  determined,  but  each 
successive  comparison  involves  an  increase  of  error 
so  that  whenever  possible  these  indirect  results  are  con- 
firmed by  a  direct  comparison  with  hydrogen  or  oxygen. 
In  the  course  of  such  a  set  of  determinations,  it  fre- 
quently occurs  that  two  or  more  entirely  different 
equivalents  are  obtained  for  the  same  element. 

This  shows  that  the  element  may  combine  in  differ- 
ent proportions  and  as  the  different  equivalents  are  in- 
variably multiples  of  the  lowest,  it  proves  that  the 
elements  combine  in  simple  and  definite  proportions 
represented  by  whole  numbers. 

To  illustrate  this :  suppose  in  the  experiment  des- 
cribed with  black  oxide  of  copper,  we  substitute  red 


4  CHEMICAL  EQUIVALENTS. 

oxide — 4  grams  as  before,  and  weigh  the  copper  remain- 
ing— 3.553  grams.  We  shall  have  to  obtain  the  equiv- 
alent of  copper 

3.553  ::  0.447  ::x:  8.     x  =  63.6, 

which  is  exactly  double  that  obtained  from  the  black 
oxide,  so  that  8  parts  by  weight  of  oxygen  combine  in 
one  case  with  31.8  parts  of  copper  and  in  the  other  with 
63.6  parts.  This  leads  to  the  conclusion  that  the  red 
oxide  contains  exactly  one  half  as  much  oxygen  as  the 
black  oxide  in  proportion  to  copper.  When  we  attempt 
to  construct  formulae  from  these  results,  we  are  con- 
fronted by  a  difficulty,  for  Cu2O  and  CuO  or  CuO  and 
Cu(X  equally  well  represent  the  proportions  of  oxygen 
to  copper  found.  To  distinguish  between  these,  both 
molecular  weight  determinations  and  study  of  the  salts 
corresponding  to  the  respective  oxides  and  the  analogous 
compounds  of  similar  elements,  are  necessary. 

Equivalents  may  also  be  determined  electrolytically, 
for  according  to  Faraday's  Law,  the  same  quantity  of 
electricity  precipitates  or  liberates  chemically  equivalent 
quantities  of  all  elements.  Suppose  we  connect  in  series, 
cells  containing  suitable  solutions  of  cuprous  copper, 
cupric  copper  and  of  silver  respectively,  each  provided 
with  platinum  electrodes;  then  pass  a  weak  current 
through  the  three,  as  they  are  connected  in  series  they 
obtain  the  same  current;  so  that  the  weights  of  metal 
precipitated  are  proportional  to  the  chemical  equivalents. 
Suppose  the  weights  precipitated  are  silver  0.756  gram ; 
copper,  0.4452  in  one  case,  and  0.2226  in  the  other.  If 
we  know  one  equivalent  we  can  find  the  others ;  if  we 
know  that  of  silver  to  be  108,  to  find  the  two  equivalents 
•of  copper  we  have 

0.756  :  0.4452  :  :  108  :  x.     x  =  63.6. 

0.756  :  0.2226:  :  108  :  y.     y  =  3i.8. 


ATOMIC   WEIGHTS.  5 

For  the  relations  between  chemical  and  electro- 
chemical equivalents  see  Chapter  X. 

ATOMIC     WEIGHTS. 

The  atomic  weight  of  an  element  is  the  weight  of 
an  atom  compared  with  the  weight  of  an  atom  of  hydro- 
gen taken  as  unity  or  an  atom  of  oxygen  taken  as  six- 
teen. This  latter  definition  is  very  generally  adopted, 
for  the  ratio  of  oxygen  to  hydrogen  has  been  found  by 
most  careful  experiments  to  be  not  16  to  I  but  15.88  to 
i  and,  as  the  value  for  oxygen  enters  into  a  great  num- 
ber of  atomic  weight  calculations,  it  is  more  convenient 
to  take  it  as  a  whole  number,  16,  and  make  it  the  base 
of  the  system.  If  we  use  O  =  16,  hydrogen  then  becomes 
1.008  when  great  exactness  is  required. 

The  atomic  weight  is  equal  to,  or  a  multiple  of,  the 
equivalent,  for  an  element  can  combine  with  one  or  more 
hydrogen  atoms  or  some  other  element  in  different  pro- 
portions depending  on  its  valence,  and  while  this  affects 
its  equivalent  it  does  not  its  atomic  weight  which  is  an 
invariable  property  of  the  element.  To  determine  which 
multiple  of  the  equivalent  is  the  atomic  weight  is  some- 
times a  matter  of  difficulty ;  which  requires  a  knowledge 
of  the  laws  and  methods  of  physical  chemistry  for  its 
certain  solution,  particularly  those  regarding  molecular 
weights,  which  are  not  within  the  province  of  this  book. 
If,  however,  we  take  the  least  common  multiple  of  the 
equivalents,  when  the  element  has  several,  we  get  a 
number  which  satisfies  all  analytical  data  and  is  the 
atomic  weight  subject  to  confirmation  by  physical  data 
and  the  periodic  law. 

In  order  not  to  leave  the  question  of  atomic  weights 
in  such  a  vague  condition,  let  us  assume  the  correctness 


ATOMIC  WEIGHTS. 


of  the  following  facts,  for  the  proof  of  which  the  student 
is  referred  to  books  on  physical  chemistry. 

1.  That  the  product  of  the  specific  heat  of  an  ele- 
ment in  the  solid  state  and  the  atomic  weight  is  a  con- 
stant (approximately)  6.4.     Law  of  Dulong  and  Petit. 

2.  That  equal  volumes  of  all  gases  under  the  same 
temperature  and   pressure   contain  the  same  number  of 
molecules.     Law  of  Avogadro. 

3.  That  the  gaseous  molecules  of  all  elements  con- 
sist of  two  atoms,  except  potassium,  sodium,  zinc,  cad- 
mium,  and  mercury  which  are  monatomic,   phosphorus 
and  arsenic  which  are  tetratomic  and  sulphur  which  is 
variable  according  to  temperature.    (Phosphorus,  arsenic 
and  sulphur  are  diatomic  however  at  very  high  temper- 
atures). 

With  the  aid  of  these  facts  we  can  readily  confirm 
our  provisional  atomic  weights  as  follows: — The  least 
common  multiple  of  our  equivalents  for  copper,  31.8  and 
63.6,  is  63.6;  the  specific  heat  of  copper  is  0.095  (see 
table)  63.6  X  0.095  =  6.042  ;  a  result  which  approximates 
6.4  with  sufficient  closeness  to  show  any  other  multiple 
of  the  equivalent  to  be  incorrect.  Again  we  found  the 
equivalent  of  silver  to  be  108 ;  its  specific  heat  is  0.057 
and  the  product  is  6.15,  a  result  which  shows  that  the 
atomic  weight  and  equivalent  of  silver  are  identical. 

This  principle  can  be  applied  to  all  elements  whose 
specific  heats  in  the  solid  state  are  accurately  known, 
but  will  not  aid  us  to  confirm  the  atomic  weight  of  many 
of  the  non-metallic  elements.  To  do  this  we  must  con- 
sider the  other  facts.  Take  a  diatomic  molecule  such 
as  chlorine  and  compare  the  weight  of  a  liter  of  this 
with  the  weight  of  a  liter  of  hydrogen,  at  the  same  tem- 
perature and  pressure.  According  to  the  law  of 
Avogadro  these  volumes  contain  the  same  number  of 


ATOMIC  WEIGHTS.  7 

molecules.  Let  x=number  of  molecules  in  a  liter,  then 
weight  x  chlorine  molecules :  weight  x  hydrogen  mole- 
cules: :  weight  one  chlorine  molecule  :  weight  one  hydro- 
gen molecule.  And  as  one  chlorine  molecule  and  one 
hydrogen  molecule  each  contain  two  atoms,  we  have, 
weight  liter  of  chlorine :  weight  liter  of  hydrogen : : 
atomic  weight  of  chlorine  :  atomic  weight  of  hydrogen : 
or  3.1673  :  0.0896: :  x  :  i.  x=35.35.  A  result  which  con- 
firms with  sufficient  accuracy  our  equivalent  35.5  as  the 
atomic  weight  of  chlorine, 

To  determine  the  atomic  weight  of  mercury. 

1.  Ignite  5  grams  of  mercuric  oxide  and  weigh  the 
resulting  mercury — 4.62    grams.     The    loss   in    weight 
0.38  gram   is  oxygen  and  the  equivalent  of  oxygen  is  8 
so  0.38  :  4.62  :  :8  :  x.    x  =  9/.26.    The  equivalent  of  mer- 
cury. 

2.  Dissolve  5  grams  of  corrosive  sublimate  in  water, 
acidify  with  nitric   acid  and  precipitate  the  chlorine  in 
combination  by  silver  nitrate.    Weigh  the  silver  chloride, 
5.2952  grams.     We  have  found  the  equivalents  of  chlor- 
ine and  silver  to  be  35.5  and  108,  therefore  to  get  the 
chlorine  in  this  precipitate  we  have 

108  +  35.5  :  35.5  :  :  5.2952  :  x.      x  =  1.3099  grams, 
the  amount  of  chlorine   combined   originally  with    the 
mercury;   5 — 1.3099  =  3.6901,   the  amount  of  mercury 
present   in   the  five  grams  of  mercuric  chloride.     So  to 
get  the  equivalent  of  mercury  we  have, 

3.6901  :  1.3099  :  :  x  :  35.5.      x  =  ioo.+ 

3.  Fuse  5  grams  of  calomel  with  an  excess  of  alka- 
line carbonate    in    a  porcelain   crucible,    leach   out  the 
alkaline  chloride  formed  and  precipitate  it  with  an  excess 
of  silver  nitrate  ;  weigh  the  silver  chloride,  3.0466  grams. 


8  ATOMIC  WEIGHTS. 

Calculate  the  chlorine  present  as  before 

H3-5  :  35-5  '•  :  3-0466  :  x.     x  =  0.7537  gram. 
To  calculate  the  equivalent  of  mercury  we  have 

5-  —  0-7537  :  0-7537  :  :  x  :  35.5.  x  =  200. 
We  have  before  us  three  equivalents  of  mercury,  between 
the  first  and  second,  the  indirect  determination  has  the 
greater  probability  of  accuracy  on  account  of  loss  of 
mercury  by  heating,  so  the  two  reliable  equivalents 
are  100  and  200.  Therefore  the  provisional  atomic 
weight,  the  least  common  multiple,  is  200.  This  is  easily 
confirmed  by  the  specific  heat :  0.0319  times  200  =  6.38. 
So  that  200  is  the  atomic  weight  of  mercury  and  we  can 
assign  to  the  two  chlorides  formulae  of  HgCl  and  HgCL 
unless  their  molecular  weights  show  them  to  be  multi- 
ples. 


EXAMPLES. 


For  specific  heat  of  metals  see  tables  on  page  162. 

1.  Five  grams  of  pure  zinc  were  treated  with  dilute 
hydrochloric  acid  and  the  hydrogen  evolved  was  dried, 
measured  and  the  corresponding  weight  calculated;  this 
was  0.1529    gram.      What  is  the  equivalent    and    the 
atomic  weight  of  zinc?  Ans.     32.7  and  65.4. 

2.  One  gram  of  pure  metallic  magnesium  was  burned 
in  an    atmosphere  of  oxygen   and  the  resulting  oxide 
collected  and  weighed.       Weight   1.658  grams.       What 
is  the  atomic  weight  of  magnesium  ?        Ans.     24.31. 

3.  a.  Two  grams  of  silver  were  'dissolved  in  nitric 
acid  and  bromine  water  added  until  no  more  precipitate 
formed.     Weight  of  silver  bromide  was   3.481    grams. 
Given  the  equivalent  of  silver,  108.    What  is  the  equiva- 
lent of  bromine  ?  Ans.     79.97. 

b.  Given  the  weight  of  a  liter  of  bromine  vapor, 
7. 1675  grams  and  the  weight  of  a  liter  of  hydrogen  under 
the  same  conditions,  0.0896  gram.  What  is  the  atomic 
weight  of  bromine  ?  Ans.  79.99. 

4.  Three  grams  of  silver  are  precipitated  by  an  ex- 
cess  of  iodine.     Weight  of  silver   iodide   was  6.5277 
grams. 

Three  grams  of  silver  are  precipitated  by  an  excess 
of  chlorine.  Weight  of  silver  chloride,  3.9861  grams. 

Given  the  equivalent  of  chlorine,  35.5.  What  is  the 
equivalent  of  iodine  ?  A  ns .  1 2  7 . 


10  EXAMPLES. 

5.  Roscoe  found   that  6.4428  grams    of  Cape  dia- 
monds gave   23.6275    grams  of  carbon    dioxide   when 
burned  in  air. 

Van  der  Plaats  found  that  30.8891  grams  of  sugar 
charcoal  gave  113.2320  grams  of  carbon  dioxide.  Given 
O  =  1 6.  What  is  the  atomic  weight  of  carbon  ? 

Ans.      12. 

6.  Richards  and  Rogers  found  that  2.35079  grams 
of  zinc  bromide  gave   3.91941  grams  of  silver  bromide. 
Given    Br  =  79.95   and  Ag  =  107.92.      What  are  the 
equivalent  and  atomic  weights  of  zinc  ? 

Ans.     32.73  and  65.46. 

7.  Winkler     converted      electrolytically     deposited 
nickel  to  a  dry  chloride  and  then  weighed   the  chlorine 
a's  silver  chloride.     He  obtained  from  0.301 1   gram  of 
nickel    1.4621    grams  of  silver  chloride.     Given  Ag  = 
107.92  and  Cl  =  35.37.     What  are  the  equivalent  and 
atomic  weights  of  nickel?  Ans.     29.5  and  59. 

8.  Deville  found  that  0.6763  gram  of   boron  chlor- 
ide, obtained  by  the  action  of  gaseous  hydrochloric  acid 
on  boron,  gave  2.4770  grams  of  silver  chloride.      Given 
Ag  =  107.92  and  Cl  =  35.37.     What  is   the    chemical 
equivalent  of  boron  ?  Ans.     3.75. 

9.  Dumas    found    that  it  required  4.168    grams  of 
silver  (as  nitrate)  to  precipitate  the  chlorine  from  4.0162 
grams  of  anhydrous  barium  chloride.     Given  the  equiva- 
lents of  silver  and  chlorine,  107.92  and  35.37.     What  is 
the  chemical  equivalent  of  barium?  Ans.     68.62. 

10.  Stas  found  that  after  adding  7.25682  grams  of 
potassium  chloride  to  10.51995  grams  of  silver,  dissolved 
in   nitric  acid,  that  0.0194  gram   of  silver  remained  in 
solution.     Given    the  atomic  weights    of  chlorine    and 
silver  as  35.37  and  107.92.    Calculate  that  of  potassium. 

Ans.     39.2. 


EXAMPLES.  11 

11.  Mallet  determined  the  atomic  weight  of  alumi- 
nium by  the  following  methods  : 

(a)  by  oxidizing  to  water  the  hydrogen  evolved  when 
aluminium    was  dissolved  by  sodium  hydroxide  :  5.2632 
grams  of  aluminium  gave  5.2562  grams  of  water. 

(b)  by  precipitating  the  bromine  in  aluminium  bro- 
mide as  silver  bromide  :  8.6492  grams  of  aluminium  bro- 
mide required  10.4897  grams  of  silver. 

Calculate  the  atomic  weight  of  aluminium.  Given 
Ag  =  107.92,  Br  =  79.95,  O  =  16,  H  =  1.008. 

Ans.     (a)  2  7. 06.     (b)  2  7. 1 05. 

12.  If  5   grams  of  a  metallic  bromide  gave  7.2552 
grams  of  iodide,  and  the  atomic  weights  of  bromine  and 
iodine  are  79.95  and   126.85   respectively.     What  would 
be  the  atomic  weight  of  the  metal  if  it  were  univalent, 
bivalent  or  trivalent  ?  Ans.     24,  48,  72. 

After  this  introduction  on  the  determination  of  atomic 
weights,  the  student  should  have  a  definite  idea  of  their 
meaning  and  how  they  can  be  obtained.  In  the  calcula- 
tions and  examples  from  this  point  the  atomic  weights 
given  on  page  159  will  be  used  unless  especially  men- 
tioned. These  were  adopted  by  the  American  Chemical 
Society  and  published  in  the  Journal  for  February,  1899. 


CHAPTER    II. 

CALCULATION  OF  FORMULAE  AND  PERCENTAGE. 
FORMULA    FROM    PERCENTAGE. 

In  any  chemical  compound,  if  we  divide  the  percent- 
ages obtained  from  analysis  by  the  atomic  weights  of 
the  elements,  we  shall  get  numbers  which  represent  the 
ratio  of  atoms,  and  on  reducing  these  to  the  nearest 
whole  numbers,  obtain  the  empirical  formula — the  sim- 
plest ratio  of  atoms.  The  correct  formula  may  however 
be  a  multiple  of  this  which  would  have  the  same  percent- 
age composition.  Therefore  a  determination  of  the 
molecular  weight  is  necessary  to  decide  what  multiple  of 
the  empirical  formula  is  the  correct  one.  For  the  meth- 
ods and  calculations  used  in  determining  molecular 
weights,  except  from  the  specific  gravity  of  a  gas,  the 
reader  is  referred  to  the  numerous  text  books  on  phys- 
ical chemistry. 

The  empirical  formula  is  usually  obtained  from  an 
analysis,  and  while  the  proper  multiple  should  be  used 
in  all  equations,  yet  the  empirical  formula  serves  many 
useful  purposes  in  inorganic  analysis,  such  as  furnishing 
data  for  normal  solutions,  factors,  etc.,  just  as  correctly 
as  the  proper  molecular  formula. 

In  organic  chemistry,  this  is  not  the  case  as  the  ex- 
istence of  many  totally  different  bodies  having  the  same 
empirical  formula,  makes  it  essential  to  use  the  molecu- 

(12) 


FORMULAE  OF  MINERALS.  13 

lar  or  better  the  graphic  formulae.  The  following  ex- 
amples illustrate  the  calculation  of  empirical  formulae : 

A  salt  gave  on  analysis  52. 45%  of  potassium  and 
47-55%  of  chlorine, 

52.45  +-39.11  =  1.341, 

47-55  -*•  35-45  =  i-34i» 

the  ratio  of  atoms  is  1.341  to  1.341  or  i  to  i,  so  the  for- 
mula has  an  equal  number  of  atoms  of  potassium  and 
chlorine  and  the  empirical  formula  is  KC1. 

A  salt  gave  on  analysis  the  following : 
Phosphorus  21. 82% -r-  31.02  =0.703!    Rado 

Hydrogen  0.71%+-  i.oi     =0.703'     of 

Sodium  32.43%  •*•  23-05  =  1-407  |  Atoms 

Oxygen  by  difference  45.02%--  16.       =2.814] 

If  we  divide  the  numbers  representing  the  ratio  of 
atoms  by  the  greatest  common-divisor  0.703  we  get  the 
ratio  expressed  in  the  nearest  whole  numbers,  or  phos- 
phorus i ;  hydrogen  i ;  sodium  2  ;  oxygen  4.  The  em- 
piiical  formula  is  therefore  PHNa2O4  or  as  it  is  usually 
written  NaoHPO4. 

Analysis  of  ignited  magnesium  ammonium  phosphate 
gave  the  following  results  : 

Magnesium  21.82%-:-  24.29  =  0.898-^0.449  =  2. 
Phosphorus  27.86% -- 31.02  =  0.898^0.449  =  2. 
Oxygen  50.32%--  16.  =  3.145-^0.449  =  7. 

The  greatest  common  divisor  of  the  numbers  repre- 
senting the  ratio  of  atoms  is  0.449  :  dividing  by  this  we 
obtain  the  nearest  whole  numbers.  So  the  empirical 
formula  is  Mg2P2O7. 

FORMULAE    OF    MINERALS. 

From  the  results  of  analysis  the  formula  of  a  mineral 
can  be  calculated  as  in  the  case  of  any  other  empirical 
formula ;  but  as  one  element  or  group  often  replaces 


14  FORMULAE  OF  MINERALS. 

another  to  a  greater  or  less  extent  without  altering  the 
mineralogical  character,  those  elements  which  may  re- 
place each  other  are  grouped  together  and  represented 
by  the  letter  R.  So  oxides  of  the  monatomic  elements, 
such  as  the  alkalies,  are  represented  by  R2O ;  oxides 
of  the  diatomic  elements,  such  as  calcium,  magnesium, 
barium,  ferrous  iron,  manganese,  etc.,  by  RO;  oxides  of 
aluminium,  chromium,  ferric  iron,  manganic  manganese, 
etc.,  by  R2O3  and  so  on.  In  this  way  the  composition 
of  the  mineral  is  indicated  and  especially  the  ratio  of 
basic  to  acid  oxides. 

For  example,  analysis  of  Franklinite  gave  the  follow- 
ing results  :* 

(0  (*)  (3) 

Fe2O3  60.52  66.34  67.42 

Mn2O3  6.79 

ZnO  19-44  20.26  6.78 

MnO  12. 81  12.31  9.53 

FeO  15.65 

We  see  in  (i)  a  replacement  of  part  of  the  Fe2O3  by 
Mn2O3 ;  so  that  R2O3  in  this  case  is  ferric  iron  and  man- 
ganic manganese,  comparing  (i),  (2),  and  (3),  we  see 
that  the  percentages  of  ZnO,  MnO  and  FeO  vary  greatly, 
these  constitute  the  RO  group. 

In  (i)  calculate  from  the  percentage  of  Mn2O3  the 
equivalent  percentage  of  Fe.3O3.  To  do  this  we  have, 
the  percentage  of  Mn2O3  is  to  the  percentage  of  Fe2O3 
as  the  molecular  weight  of  Mn2O3  is  to  the  molecular 
weight  of  Fe2O3.  The  molecular  weight  of  Mn2O3  is 
(2  x  55)  +  t  (3  x  1 6)  or  158  and  that  of  Fe2O3  is  (2  x  56) 
+  (3  x  1 6)  or  1 60.  So  the  proportion  becomes, 
6.79  :  x  ::  158  :  160.  x  =  6.88%. 

*  Dana's  Mineralogy. 

f  In  these  calculations  the  atomic  weights  to  one  decimal  place  are  sufficient. 


FORMULAE  OF  MINERALS.  15 

In  the  same  way,  using  the  proper  molecular  weights, 
calculate  the  percentage  of  MnO,  to  the  equivalent  per- 
centage of  ZnO, 

12. 81  :  y  :  :  71.  :  81.4.     y  =  14.68%. 

We  have  now  Fe2O3,  60.52% +  6.88%  ==  67.40%  and 
ZnO,  19.44%  +  H-68%  =  34-12%.  Dividing  these  by 
the  molecular  weights  we  get 

67.40%^-  160   =  0.421  -r-  0.420  =  i.  -f- 
34.12%-r-  81.4  =  0.419-^-0.420  =  i.  — 

These  results  are  almost  identical,  so  the  ratio  of  R2O3 
to  RO  is  i  to  i  and  the  formula  is  represented  by 
RO,R2O3. 

There  was  no  particular  reason  for  calculating  the 
percentage  of  MnO  to  ZnO  rather  than  ZnO  to  MnO ; 
the  result  when  divided  by  the  proper  molecular  weight 
will  be  the  same  in  either  case. 

The  typical  formula  can  also  be  found  by  dividing 
the  percentage  of  each  oxide  by  its  molecular  weight 
and  then  grouping  together  the  ratios  as  follows — 

(2)     Fe2O3  66.34-^  160    =0.414. 

ZnO     20.26-7-81.4  =  0.249)    _ 
MnO    12.31  -r-    71.  =  0.172  j 

The  ratio  is  approximately  one  to  one  as  before.  The 
lack  of  agreement  in  the  figures  representing  the  ratio 
is  probably  due  to  a  deficiency  in  the  percentage  of  R2O3 
elements  for  the  percentages  only  add  up  to  98.91. 

From  the  third  analysis  we  get  0.421  :  0.435,  again 
showing  the  ratio  to  be  one  to  one. 

So  from  the  results  of  these  three  analyses  we  find  the 
formula  of  Franklinite  to  be  (Zn  Mn  Fe)O,  (Fe  Mn)2O3. 


16  FORMULAE  OF  MINERALS. 

Dana's  Mineralogy  gives  the  following  analyses  of 
garnets, 

(0  (*) 

SiO2  40.90                 39.85 

A12O3  22.81                  22.07 

Fe2O3  1.13 

Cr203  1.48 

FeO  13.34 

MnO  0.38 

MgO  16.43                   0.68 

CaO  4.70                 36.31 

The  RoO3  bases  are  calculated  to  ALO:,  as  the  per- 
centages of  Fe2O3  and  Cr2O:J  are  small  and  indicate  a  par- 
tial replacement  of  the  A1,O,.  With  the  KG  bases  we 
may  calculate  either  to  CaO  or  MgO  and  divide  the  sum 
of  the  resulting  percentages  by  the  molecular  weight  of 
the  oxide  we  select.  After  calculating  Cr2O3  and  Fe2O:J 
to  AL>O3  and  FeO,  MnO,  and  MgO  to  CaO  we  get 

(0 

SiOo  40.90%  —    60.4=0.677.     3. 

A12O3(R2O3)  23.80%—  102.  2  =  0.236.     i. 

CaO(RO)  38-37%-   56.    =0.685.     3. 


SiO2  39-85%  -s-    60.4  =  0.660.     3. 

A1203(R208)          22.79%-  102.2  =  0.223.     i. 
CaO  (RO)  37-26%^    56.     =0.665.     3. 

So  in  both  cases  the  type  formula  is  3RO,  R2O3,  3SiO2. 
In  connection  with  silicates  both  natural  and  artificial 
(slags)  the  term  "  oxygen  ratio  "  is  often  used.  This 
means  the  ratio  of  the  oxygen  atoms  belonging  to  the 
basic  groups  to  those  combined  with  silicon.  In  the 
garnet  we  have  six  oxygen  atoms  combined  as  basic 
oxides  and  six  as  an  acid  oxide  (SiO2),  hence  the  oxygen 


PERCENTAGE  COMPOSITION  FROM  FORMULA.  17 

ratio  is  one  to  one,  or  if  it  is  desired  to  distinguish  be- 
tween the  basic  oxides  1:1:2.  In  writing  these  ratios 
the  order  adopted  is  first  RO,  then  R»O3  and  last  SiO2. 

PERCENTAGE    COMPOSITION     FROM     FORMULA. 

If  we  have  the  formula  of  a  compound  whether  mole- 
cular or  empirical  and  the  atomic  weights  of  its  elements 
the  calculation  of  the  percentage  composition  is  made  as 
follows. — Take  for  example  sodium  chloride ;  we  have 
the  formula  NaCl  and  the  atomic  weights  Na  23.05  and 
Cl   35.45.       The  molecular  weight  is  the  sum  of  the 
atomic    weights,  23.05  +  35-45   =   58.5,  the   molecular 
weight  of  NaCl.     Let  x  =  %  Na  and  y  =  %  Cl  then 
23.05  :  58.5  :  :  x  :  100.    x  =  39.402%. 
35-45  :  58.5  =  :  y  =  100.    y  =  60.598%. 
To  prove  x  +  y  =  100.00%. 

To  calculate   the  percentage    composition    of  ethyl 
alcohol,  CoH5OH.   The  molecular  weight  is  24.00+  6.05 
+  16  =  46.05.      Let  x  =  %C,  y  =  %H  and  z  =  %O. 
24.     :  46.05  :  :  x  :  100.    x  =  52.117%  C. 
6.05  :  46.05  :  :  y:  100.    y  =  13.138%  H. 
1 6.       :  46.05  :  :  z  :  100.    z  =  34-745%  Q- 

To  prove      x  -f-  y  +  z  =  i  oo  %. 
By  similar  proportions  the  percentage    composition 
of    any    substance    whose    formula    is    known    can  be 
calculated. 

CALCULATION    OF    PERCENTAGE    TO    THE    DRY    BASIS. 

Ores  and  other  material  often  contain  moisture:  it  is 
frequently  necessary  to  calculate  what  the  percentage  of 
a  constituent  would  be  if  this  water  were  removed  and 
the  reverse. 


18       CALCULATION  OF  PERCENTAGE  TO  THE  DRY  BASIS. 

If  an  ore  contains  10%  of  moisture  and  42%  of 
manganese,  what  will  be  the  percentage  of  manganese 
on  the  dry  basis  ?  Let  us  represent  the  weight  of  the  wet 
sample  by  100  parts;  then  after  drying  we  will  have 
100 — 10  (the  parts  of  water)  :  let  x  =  the  percentage  of 
manganese  on  the  dry  basis ;  then,  as  no  manganese  is  lost 
or  gained  by  the  removal  of  water,  we  have  42  X  100  =  x 
(100-10).  x  =  46.66%  or,  in  general  if  we  divide  the 
percentage  by  100  minus  the  percentage  of  moisture  we 
get  the  percentage  on  the  dry  basis. 

If  an  ore  contains  58%  of  iron  on  the  dry  basis,  what 
is  the  percentage  of  iron  when  1 8%  of  water  is  present? 

58  :  x  :  :  100  :  82.     x  =  47.56 

for  the  original  ore  may  be  regarded  as  made  up  of  82 
parts  of  dry  ore  containing  58%  of  iron  and  18  parts  of 
water  containing  no  iron. 


EXAMPLES. 


Calculate  the  empirical  formulae  from   the  following 
analytical  results. 

1.  Lead  68.30%.   Sulphur  10.55%.   Oxygen  by  dif- 
ference, 21.15%.  Ans.     PbSO4. 

2.  Arsenic  41.32%.  Chlorine  58.66%. 

Ans.    AsCl3. 

3.  Potassium   35.56%.  Iron   17.00%.   Cyanogen  by 
difference,  47.44%.  Ans.     K3Fe  C6N6. 

4.  Carbon  37.5%.    Hydrogen  12.5%.     Oxygen  by 
difference,  50.%.  Ans.     CH4O. 

5.  Carbon    40.67%.     Hydrogen    8.47%.    Nitrogen 
23-73%-     Oxygen  by  difference,   27.13%. 

Ans.     C2H6NO. 

6.  Carbon    46.75%.     Hydrogen    6.48%.     Nitrogen 
36.36%.  Oxygen  by  difference,    10.41%. 

Ans.     C6H10N40. 
Calculate  the  formulae  of  the  following  minerals. 

7.  Pyrargyrite.     (*)      Sulphur     17.81%.  Antimony 
22.45%.     Silver  59-75%-  Ans.     Ag3SbS3. 

8.  Enargite.     Sulphur    32.69%.     Arsenic   19.47%. 
Copper  47.84%.  Ans.     Cu3AsS4. 

9.  Bournonite.    Sulphur  19.36%.  Antimony  23.57%. 
Arsenic  0.47%.    Lead  41.95%.     Copper  13.27%.    Iron 
0.68%.  Ans.     (PbCu2)3Sb2S6. 

10.  Atacamite.     Chlorine  16.45%.     Copper  14.72%. 
Copper  oxide  (CuO)  55.26%.     Water  13.57%. 

Ans.     Cu2Cl  H3O3. 

*  These  percentages  are  taken  from  Dana's  Mineralogy. 
(19) 


^0  EXAMPLES. 

11.  Barytocalcite.     Carbonic  acid  29.44%.     Barium 
oxide    50.36%.      Calcium    oxide    19.22%     Manganous 
oxide  0.25%.  Ans.     BaCO3,CaCO3  or  RCO3. 

12.  Given    the    following  analysis  of  augite:    SiOo 
54.28%,  AL03o.5i%,  Fe203o.98%  FeO  i.9i%,MgO~ 
17.30%,  CaO   25.04%.      Calculate  its  formula  and  oxy- 
gen ratio.  Ans.     CaMgSioO6  and  i  :  2. 

13.  Find  the  type  formula  and  oxygen    ratio    for    a 
mineral    whose    composition    is    SiO2    38.03%,    AloO.{ 
20.83%,  FeO  36.15%.  MnO  2.14%,  MgO  0.97%,  CaO 
2.73%.  Ans.     3RO,  R,O,,  3SiO2  and  i  :  i. 

14.  The   mineral    Zirkelite  on  analysis  yielded    the 
following  results:     ZrO2  52.89%,  Ti(X  14.95%,   ThO.,- 
7.31%,"  Ce303  2.52%,  (Y203)?  0.21%,  U02i.4o%,  FeO 
7.72%   CaO    10.79%,   MgO   0.22%,  loss    on    ignition 
1.02%.     What  is  the  formula  ?  (*) 

Ans.     RO.2,(Zr,  Ti,  Th,)O,,. 

Calculate  the  percentage  composition  of  the  following 
compounds. 

15.  Potassium  Sulphate,  K2SO4. 

Ans.     K,  44-88%.    S,  18.40%.     0,36.72%. 

1 6.  Manganous  Sulphate,  MnSO4. 

Ans.     Mn,  36.42%.    S,  21.23%.    O,  42.36%. 

17.  Lithium  Phosphate,  Li3(PO4)2. 

Ans.     Li,  9.99%.    P,  29.38%.    0,60.62%. 

1 8.  Zinc  pyrophosphate  Zn2P2O7. 

Ans.     Zn,  42.91%.    P,  20.35%.    0,36.72%. 

19.  Urea,  CH4N2O  and  Ammonium  Cyanate,  NH4 
CNO. 

Ans.     C,  19.96%.  H,6.7i%.  ^46.71%.  0,26.61%. 

20.  Benzyl  Chloride  C6H6CH2C1. 

Ans.     C,66.4o%.    H,5.58%.    C1,28.O2%. 


Mineralogical  Magazine  1895-97.  XI,    180. 


EXAMPLES.  21 

21.  Nitrotoluene  CBH4(NO3)  CH3. 

Ans.    C,  61.27%.       H,  5.14%.       N,    10.24%.     O, 

23-34%. 

22.  Indigo  C16H10NaO2. 

Ans.    C,  73-24%   H,  3.84%.  N,  10.71%.  O,  12.20%. 

23.  Potassium  ferrocyanide,  crystallized,  K4Fe(CN)6, 
3H20.         Ans.    K,  37.00%.    Fe,  13.25%.    €,17.03%. 

N,  19.92%.    H,  1.43%.    O,  11.35%. 

24.  A  sample  of  ore  contained  20%  of  lead,  14%  of 
zinc  and  14%  of  water.    What  are  the  percentages  of 
lead  and  zinc  on  the  dry  basis. 

Ans.     Lead  23.25%,  Zinc  16.28%. 

25.  A   clay  was  partially  dried   and  then  contained 
SiO.,  50%,  H2O   7%;    the   original    sample   contained 
1 2%  of  water.     What  is  the  percentage  of  SiO3  in  the 
original  sample.  Ans.     47.37%. 


CHAPTER   III. 

CALCULATIONS    OF    MIXTURES    HAVING   A    COMMON 
CONSTITUENT. 

To  calculate  the  weights  of  chlorine  and  bromine  in 
a  mixture  of  silver  chloride  and  bromide. 

If  we  have  a  solution  containing  chlorides  and  bro- 
mides and  wish  to  determine  the  weights  of  chlorine  and 
bromine  present,  we  can  precipitate  these  elements  by 
silver  nitrate  and  weigh  the  resulting  mixture  of  AgCl 
and  AgBr,  weight  1.50  grams.  Then  by  scorification  or 
some  other  method,  determine  the  silver;  weight  i.oo 
gram.  We  have  the  atomic  weights  of  silver  107.92, 
chlorine  35.45,  bromine  79.95,  the  weight  of  the  mix- 
ture and  the  weight  of  the  common  constituent,  silver. 

If  all  the  silver  were  present  as  chloride   the   weight 
would  be   1.32848  grams:  for  as  we  know  the  formula 
and  atomic  weights  we  have  the  proportion 
107.92  :    143.37    ::        I.     :     x. 

At.  wt.  Ag :  Mol,  wt.  AgCl : :  wt.  Ag:  wt  AgCl,  which 
gives  x  =  1.32848,  the  weight  of  silver  chloride  corre- 
sponding to  the  silver  found.  But  we  have  1.50  grams  of 
mixed  chloride  and  bromide,  hence  1.50-1.32848  or 
0.17152  gram  is  the  excess  in  weight  due  to  the  greater 
atomic  weight  of  bromine  and  therefore  proportional  to 
the  weight  of  bromine  present.  So  to  obtain  the  weight 
of  bromine  we  have  : 

79-95      -    35-45      •      79-95      "      0.1715        :  x. 
At.  wt.  Br  -  At.  wt.  Cl :  At.  wt.  Br  : :  excess  in  wt. :  wt.  Br. 

(22) 


MIXTURES  HAVING  A  COMMON  CONSTITUENT.  23 

x  =  0.30815,  weight  Br.    The  weight  of  chlorine  will  be  : 
Weight  silver,  chlorine  and  bromine  1.50000  grams 

weight  silver  (i.)  and  bromine  (0.30815)  1.30815  grams 

or  0.19185  gram 

To  calculate  the  chlorine  directly :   find  the  weight  of 
silver  bromide  corresponding  to  the  silver  present. 
107.92  :  187.87  :  :    i  :  x.      x  =  1.74082  grams. 
But  we  have  only  1.5  grams  of  the  mixture  so  1.74082  - 1.5 
or    0.24082  is   the  deficiency  due   to  the  lower  atomic 
weight  of  chlorine  and   proportional    to    the   weight  of 
chlorine  present. 
Let  y  =  weight  of  chlorine,  then  we  have 

79-95—35-45  :  35-45  •  :  0.2408  :  x 

and  x  =  0.19183  the  weight  of  chlorine   present,  which 
confirms  the  result  obtained  by  difference. 

It  would  be  more  logical  to  say  that  the  excess  in 
weight  is  due  to  the  greater  molecular  weight  of  AgBr 
or  the  deficiency  due  to  the  lower  molecular  weight  of 
AgCl,  but  this  only  means  the  addition  of  107.92  in  both 
cases  which  does  not  affect  the  result  after  subtraction. 

Perhaps  the  following  method  is  more  easily  compre- 
hended from  a  mathematical  standpoint  but  it  is  longer 
and  the  use  of  logarithms  is  recommended. 

Let   x  =  weight  of  AgCl  and  y  =  weight  of  AgBr 
then 

1 07.02 

—  x   =  weight  of  silver  in  AgCl  and 
H3-37 

-~~—  y  =  weight  of  silver  in  AgBr ;  so,  using  the  same 

157.57 

weights  as  before  we  have  the  two  equations, 

(1)  x  -f-  y  =  1.50  and 

(2)  IQ7-92    x  +     107-9.2      =  j 


24  MIXTURES  HAVING  A  COMMON  CONSTITUENT. 

reducing  to  decimals  we  have  o.  75274  x  +  0.57444  y  =  I 
substituting  for  x  its  value  in  terms  of  y  from  equation, 
(i)  we  have 

0.75274  (1.5-  y)  +  0.57444  =  i      Solving 
0.1291  =  0.1783  y.     y  =  0.7241,  wt.  AgBr. 

substituting  this  value  in  equation  (i)  we  get  x  =  0.7759 
wt.  AgCl ;  from  which  by  the  following  proportions, 

H3-37  :  35-45  :  •  Q-7759  •  wt.  Cl  and 
187.87  :  79.95  :  :  0.7241  :  wt.  Br; 

we  get  wt.  Br  0.3081  and  wt.  Cl  0.1918  gram. 

Suppose  we  have  the  weight  of  a  mixture  of  potas- 
sium and  sodium  sulphates,  0.3710  gram,  and  the  weight 
of  SO3,  0.200  gram  ;  to  calculate  the  weights  of  K2O 
and  Na2O.  Calculate  the  SO:5  all  to  Na2SO4, 

80.07  :  H2-1?  :  :  0.2  :  x-  x  —  °-355I- 
but  as  we  have  0.3710  gram  of  mixed  sulphates,  the 
excess  in  weight  0.0159  gram  is  due  to  the  excess  in 
molecular  weight  of  KoO  over  Na»O  and  is  therefore 
proportional  to  the  weight  of  K2O,  so  we  have  the  pro- 
portion, 

32.12  :  94.22  :  :  0.0159  :  x.     x  =  0.0466 
K2O  0.0466  gram.  KoO.-f  NaoO  +  SO3  =  0.37 10  gram, 
SO3  0.2000  gram.  K2O.  +  SO3  =  0.2466  gram 

0.2466  gram.  Na2O  —  0.1244  gram 

To  solve  this  example  by  the  other  method,  let  x  = 
weight  of  K2SO4  and  y  =  weight  of  Na2SO4,  then  x  +  y 
=  0.371. 

The  molecular  weight  of  SO3  is  80.07  an(i  tnat  °f 
K2SO4  is  174.29;  therefore 

~~        x  represents  the  weight  of  SO:!  present  as  K2SO4 


MIXTURES  CONTAINING  A  COMMON  CONSTITUENT.         25 

and  similarly  —  —  y   the   weight   of   SO:-  present  as 


^  so  we  have  for  a  second  equation, 

°'   '   x  4-  —  '—  -v  =  0.2   or  after  multiplying1  and  re- 
174.29'        142-17 
moving  unnecessary  fractions, 

11383.5  x+  13955-4  y  =  4955-7 

substituting  from  first  equation  x  =  0.371  —  y  we  have 
1  1383-5    (o-37i  —  Y)  +  13955-4  y  =  4955-7  or 
257I-9  y  =  732-4-  y  =  0-2847,  weight  of  Na2SO4; 
and  as  x  +  y   =0.3710.  x   =   0.0863,  weight  K2SO4. 
From    which    the  following    proportions  will  give  the 
weights  of  K3O  and  Na2O 

94.2  :  174.29  ::  wt.  K2O  :  0.0863.  wt.  K2O  ==  0.0466. 
62.1  :  142.17:  :  wt.  Na2O  :  0.2847.  wt.  Na2O  =  0.1243. 

The  following  is  a  more  practical  example.  One 
gram  of  commercial  sodium  bicarbonate  gave  CO.,,, 
0.5097  gram  and  Na2O,  0.3681  gram.  Calculate  the 
weights  of  NaHCO:5  and  Na2CO3.  Let  x  be  the  weight 
of  Na2CO.^and  y  the  weight  NaHCO:5;  then, 

(i)   44.01    44.01 

?-£—  x  +  ~  -  y  =  0.5097  and 
1  06.  1  1    84.07  * 


(2)        62.1         ,     62.1 


106.11          168.14 


y  =  0.3681 


The  molecular  weight  of  NaHCO3  is  doubled  in  the 
second  equation  as  it  takes  two  molecules  to  give  one 
Na-jO.  Reduce  the  fractional  coefficients  to  the  nearest 
decimals, 

(1)  0.4147  x +  0.5236  y  =  0.5097 

(2)  0.5852  x  -f  0.3693  y  =  0.3681 


26          MIXTURES  CONTAINING  A  COMMON  CONSTITUENT. 

substituting  for  x  in   the  second  equation  its    value  in 
terms  of  y  we  get 

8      (0.5097-0-5236  y)  +  0  3693     =  0  368l 

0.4147 

0.7192     —  0.7388  y  +0.3693  y  =  0.3681 

o-3695  Y  =  °-35  :  *•     Y  =  o-9502.  weight  of  NaHCO3. 


Substituting  for  y  in  equation  (i)  0.9502  we  get 
0.4147  x  +  0.5236  (0.9502)  =  0.5097  which  gives  x  = 
0.0294,  weight  of  Na2CO;{. 

This  problem  may  also  be  done  as  follows  :  Calcu- 
late all  the  CO2  to  NaHCO3,  according  to  the  proportion. 

44.01  :  84.07  :  :  0.5097  :  wt.  NaHCO3  : 
this  gives  0.9736  gram  ;  next  calculate  the  weight  of 
Na3O  necessary  to  combine  with  this  to  form  the  0.9736 
gram  of  NaHCO3.  This  is  0.35955  gram>  but  we  have 
present  0.3681  gram,  which  leaves  an  excess  of  0.00855 
gram.  This  must  be  employed  to  make  as  many 
NaHCO3  molecules  into  Na2CO3  molecules  as  possible 
and,  as  one  Na2O  furnishes  the  sodium  oxide  necessary 
to  convert  2NaHCO3  into  2Na2CO3,  to  find  the  amount 
of  Na2CO3  required  by  the  excess  of  Na2O  we  have  :  — 

62.1  :  212.22  :   : 

Mol.    wt.    Na2O   :    2    (Mol.    wt.    N^CO^   :    : 
0.00855          :  x 

wt.  Na2O  in  excess  :  wt.  Na2CO3 
x  =  0.0292,  weight  Na2CO3. 

The  weight  of  Na2O  present  as  Na^COg  is  twice  the 
excess  found  or  0.0171  gram.  The  rest  of  the  Na2O> 
0.3681  —  0.0171  or  0.3510  gram  is  present  as  NaHCO3. 
To  find  this  weight  :  — 

0.3510  :  x  ::  62.1  :  168.12.  x  =  0.9504,  weight  of 
NaHCO* 


FORMATION  OF  MIXTURES  OF  DEFINITE  COMPOSITION.    27 

These  examples  show  the  results  to  be  the  same  by 
•either  method,  the  figures  should  be  carried  out  to  four 
decimals  or  at  least  three  significant  figures  so  that  no 
appreciable  error  in  calculation  shall  be  added  to  the  un- 
avoidable errors  of  analysis. 

FORMATION     OF    MIXTURES     OF     DEFINITE    COMPOSITION. 

Suppose  we  wish  to  form  100  Ibs.  of  a  mixture  con- 
taining 10%  of  calcium  carbonate  and  we  have  at  our 
disposal  lots  containing  7%  and  23%  respectively. 
How  many  pounds  of  each  must  be  taken  ? 

One  lot  contains  3%  less  than  the  desired  mixture, 
the  other  13%  more,  so  if  we  take  3  parts  of  the  richer 
to  1 3  of  the  poorer  we  shall  have  a  mixture  containing 
10%. 

3  :   16  :  :  x  :  100.      x  =  18.75  Ibs.  (23%)- 
13  :  16  :  :  y  :  100.     y  =  81.25  lbs-  (?%)• 

When  the  proportions  cannot  be  seen  by  inspection 
the  following  method  can  be  used  : 

Given  two  iron  ores,  one  containing  0.42%  of  phos- 
phorus and  the  other  o.  1 7%.  How  much  of  each  must 
be  mixed  to  give  ten  tons  containing  o.  20%  of  phos- 
phorus? Let  x  be  the  weight  in  tons  of  the  0.42%  ore 
and  10  —  x  the  weight  of  the  0.17%  ore ;  then, 

x  (0.0042)  -+-  (to — x)  0.0017  =  10  (0.002). 

This  equation  is  based  on  the  equality  in  weight  of  phos- 
phorus in  the  sum  of  the  two  portions,  and  in  the  mixture. 

Solving,  we  get  x  =  1.2  tons,  10 — x  =8.8  tons. 

Given  waste  mixed  acid  from  nitrating  whose  com- 
position is  HoSO4  62.178%,  HNO:}  19.066%,  H2O 
18.756%;  sulphuric  acid  containing  97%  H2SO4;  nitric 
acid  containing  87%  HNO:j.  What  weights  of  each  must 


28    FORMATION  OF  MIXTURES  OF  DEFINITE  COMPOSITION. 

be  taken  to  give  1000  Ibs.  of  a  mixture  containing 
61%  H2SO4,22%  HNO.,  and  17%  H2O  without  adding 
water  ? 

Let  x  be  the  weight  of  waste  acid, 
y          "         '         "  sulphuric  acid, 
z          "         "        "  nitric  acid, 

then  x  (0.6218)  is  the  weight  of  H2SO4  in  the  waste 
acid,  y  (0.97)  the  weight  of  H8SO4  in  the  sulphuric  acid, 
z  (0.13)  the  weight  of  water  in  the  nitric  acid  and  so  on. 
looo  Ibs.  of  the  desired  mixture  must  contain  610  Ibs. 
H^SO^  220  Ibs.  HNO3  and  170  Ibs.  of  water.  We  have 
therefore  the  following  equations  based  on  equality  of 
weight: 

(i)  x  (0.6218)  +  y  (0.97)  =  610. 

(2)  X   (0.19066)  +  Z   (0.87)    =    220. 

(3)  x  (0.18756)  +  y  (0.03)  +  /  (0.13)  =  170. 


y  = 

0.97 

2  2O  -  X     (o.  1  9066)  /  x 

z  =  -  i  -  "  -  i  —  252.7  —  x  (0.2191) 
0.87 

substituting  in  equation  (3) 

0.18756  x  +  18.864  —  0.01923  x-h  32.851  —  0.02848  x 
=  170. 

0.13985  x  =  118.285.     x  =  845.68. 
substituting  in  equation  (i) 

y  =  610  —  (845.68x0.6218).    y  =  86  ^ 

0.97 
substituting  in  equation  (2) 

-  220—  (845.68  x  o.  19066).  6 

0.87 
Proof  845.68  +  86.77  +  67.54  =  999-99- 


EXAMPLES. 


1.  Given    a   mixture    of    KC1    and    NaCl  :   weight 
T-33°6    grams;     and    weight    of  chlorine   0.709  gram. 
What  are  the  weights  of  K  and  Na? 

Ans.      K  0.3911  gram.      Na,  0.2305  gram. 

2.  Given   a  mixture  of  PbSO4  and  BaSO4 ;    weight 
4  grams;  and  weight  of  SO4    1.62  grams.     What  are 
the  weights  of  PbSO4  and  BaSO4  ? 

Ans.     PbSO4  0.2729  gram.      BaSO4  3.7271  grams. 

3.  Given    a    mixture    of  HgCL   and    HgCl  weight 
2.35    grams    and    chlorine    0.45    gram.     What    is  the 
\\eight  of  each  chloride  ? 

Ans.      HgCl  1.4847  grams.      HgCL  o  8653  gram. 

4.  Given  a  mixture  of  Agl  and  AgBr  ;     weight  4.22 
grams  and  weight  of  silver  2. 1 1  grams.      What  are  the 
weights  of  iodine  and  bromine  ? 

Ans.     Iodine  1.4790  grams.      Bromine  0.6309  gram. 

5.  Given    a    mixture    of   NaHSO4    and    KHSO4  : 
weight  0.5  gram  and  weight   SO4  0.395  gram.     WThat 
are  the  weights  of  K  and  Na? 

Ans.     K  0.0148  gram.      Na  0.0860  gram. 

6.  Given  a  mixture  of  K2CO3  and  KHCO..;  weight 
i   gram    and    weight    C(X   0.40    gram.     What    is    the 
weight  of  each  carbonate  ? 

Ans.      KaCO3  0.3259  gram.      KHCO:!  0.6741  gram. 

7.  Given  a  mixture  of  K..C4H4O6  and  KHC4H4OC  : 
\ /eight    i    gram   and    weight  of  K  0.30  gram.      What 
j  ;  the  \veiht  of  each  tartrate  ? 


30  EXAMPLES. 

Ans.     K2C4H4O6  0.6683  gram.     KHC4H4O6  0.3316' 
gram. 

8.  Given    a    sample    of    pure    dolomite,     MgCO3 
and  CaCO3 :    weight    i    gram    and    CO2   weight   0.48 
gram.     Find  the  weights  of  CaO  and  MgO. 

Ans.     CaO  0.2862  gram.  MgO  0.2338  gram. 

9.  Given  a  mixture  of  CaH4  (PO4)2  andCa3  (PO4)2  : 
weight   i   gram  and  PO4  weight  0.70    gram.      Calcu- 
late the  weight  of  each  phosphate. 

Ans  <  CaH4(PO4);,  0.4395  gram. 
'  I  Ca3  (PO4)2  0.5605  gram. 

10.  Given  a  sample  of  commercial  potassium  bicar- 
bonate   containing   0.4679    gram   of   K2O  and    0.4338 
grain     of  CO2  in   one  gram.     How  much  KHCO3  and 
K2CO3  is  there  present? 

Ans.     KHCO3  0.98  gram.     K2CO;!  o.oi  gram. 

11.  One  gram  of  manganese  ore   gave  on  analysis,, 
manganese  0.25  gram  and  available  oxygen  0.05  gram. 
How  much  MnsO3   and   MnO2  is   there   present?     By 
available  oxygen  is  meant  the  amount  over  that  neces- 
sary to  form  MnO. 

Let  x  —  wt.  Mn2O3  and  y  =  wt.  MnO2  then  we  have 
'0998  x  + 54-99     =  Ans      Mn.,03o.2245. 

157.98  86*99 


o          o 
157.98          86.99' 

12.  Given  a  sample  of  argol  (crude  potassium 
bitartrate)  ;  one  gram  of  which  gave  0.2336  gram  of 
K2Oando73io  gram  H.,C4H4O6.  Supposing  that  there 
is  no  other  tartrate  or  potassium  salt  present,  how 
much  KHC4H4O,  and  K2C4H4O6  is  there? 

Let  x  -  wt.  K,C4H4O6  and  y  =  KHC4H4O(5,  then 
we  have 


EXAMPLES.  31 


94.22   x+    47.11  6and    15<>£ 

226.29          I88.I9  226.29         I88.I9 

07H  Ans  $a8977    gram  KHC4H4O« 

0.  73  *  •**  '*•»  •     S  T  '     ^  •«     T  T     /"V 

(  0.02       gram  KsC4H4Oa 
13.   Given    manganese    ores    containing    47% 


manganese  respectively.      How    many  pounds 


of  each  must  be  mixed  to  give  a  ton  containing  39%  of 


,        {    666.67  Ibs.  23%  ore. 
manganese.  Ans.  \  '    ,,        °<(l 

<  1333-33    Ibs.  47%  ore. 

14.  Suppose   we  have  ten   tons   of  coal    containing 
3-5%  °f  sulphur.     We  have  also  supplies  of  coal  con- 
taining 1.1%  and  1.6%  of  sulphur.      How  much  of  each 
shall  we  add  to  the  ten  tons  to  make  20  tons  containing 
2.5%  of  sulphur?     Ans.     2  tons  1.1%  and  8  tons  1.6%. 

15.  Given   iron  ores   containing  54%  of  iron    with 
0.60%  of  phosphorus  and  47%  of  iron  with  0.20%  of  phos- 
phorus.    What  is  the  percentage  of  iron  in  a  mixture  of 
these  containing  0.28%  of  phosphorus?     Ans.     484%. 

1 6.  Given  two  alloys:    a  brass  containing    copper 
65%,  zinc  35%  and  a  German  silver  containing  copper 
56%,  zinc  24%,  nickel  20%.     What  is  the  composition 
of  the  alloy  formed  by  melting  these  together  in  such  pro- 
portion that  it  contains  zinc  30%  ?    (Supposing  that  there 
is  no  loss  by  oxidation  or  volatilization  during  fusion). 

Ans.     Copper  60.90%,  zinc  30%,  nickel  9.09%. 

17.  Suppose   we  wish   to  form   100   Ibs.  of  an  alloy 
containing  copper  70%,  zinc  20%,  and  tin  10%  and  we 
have  besides  pure  copper,  a  brass,  copper  %',  zinc  ^  and 
solder,  zinc  ^,  tin  l/2.    How  much  of  each  must  be  taken? 

Ans.     Solder  20  Ibs.,  brass  30  Ibs.,  copper  50  Ibs. 

1 8.  Given  pure  silver  and  an  alloy  containing  890  parts 
of  silver  and   no  parts  of  copper.      How  many  ounces 
of  each  must  be  taken  to  form  1000  ounces  of  an  alloy, 
925  parts  of  silver  and  75  parts  of  copper? 

Ans.      681.8  ozs.  of  alloy,  318.2  ozs.  of  silver. 


CHAPTER  IV. 

CALCULATIONS  FROM   EQUATIONS. 

Chemical  reactions  are  best  represented  by  equa- 
tions ;  these  are  the  expression  of  definite  and  exact 
quantitative  changes  and  therefore  are  the  basis  of  most 
analytical  calculations. 

Take  for  example  the  simple  reaction 

BaCL  +  HSSO4  -  BaSO4  +  2HC1, 

from  a  qualitative  standpoint  this  means  that,  when  bar- 
ium chloride  is  added  to  sulphuric  acid  or  the  reverse, 
barium  sulphate  and  hydrochloric  acid  are  formed.  This 
is  not  all  that  is  shown  by  the  reaction.  It  shows  that 
one  molecule  of  barium  chloride  acting  on  one  molecule 
of  sulphuric  acid  forms  one  molecule  of  barium  sulphate 
and  two  molecules  of  hydrochloric  acid.  Now  as  the 
relative  weights  of  the  different  atoms  composing  the 
molecules  are  known,  based  on  oxygen  =  16,  we  can 
by  adding  these  together  get  the  relative  weights  of  the 
molecules  ;  so  the  reaction  becomes, 

BaCL  +  H2SO4  =  BaSO4  +  2HC1. 
13743  +  2(35-45)  4-  2  +  32-07  +  64  = 

13743  +  32.07  -f  64  + 

2(3545  +  0  or  208.33  +  98.07  =  233.5  +  72-9; 
which  expresses  a  definite  relation  by  weight.  We  do 
not  know  the  absolute  weight  of  a  molecule  but  we  do 
know  very  accurately  their  relative  weights  ;  so  that  these 
proportions  are  true  when  expressed  in  parts  by  weight ; 
and,  whatever  the  unit  of  the  system  may  be,  grains, 

(32) 


CALCULATIONS  FROM   EQUATIONS.  33 

grams,  pounds,  or  tons,  we  shall  always  have  233.5  parts 
of  barium  sulphate  and  72.9  parts  of  hydrochloric  acid 
formed  by  the  action  of  208.33  parts  of  barium  chloride 
on  98.07  parts  of  sulphuric  acid. 

Besides  showing  the  products  of  the  reaction  and  the 
relations  by  weight,  (he  equation  also  represents  relations 
of  energy  and  heat  absorbed  or  evolved.  These  will  be 
discussed  so  far  as  they  relate  to  analytical  calculations 
in  subsequent  chapters. 

Suppose  the  percentage  of  SO3  is  desired  in  magne- 
sium sulphate.  The  salt  is  dissolved,  barium  chloride 
added  and  the  resulting  barium  sulphate  weighed.  Ac- 
cording to  the  preceding  reaction,  for  every  molecule  of 
SO3  present  there  will  be  a  molecule  of  BaSO4  precipi- 
tated; hence  to  find  the  weight  of  SO3,  we  have:  mole- 
cular weight  of  BaSO4:  molecular  weight  of  SO3  :  : 
actual  weight  of  BaSO4:  corresponding  weight  of  SO3, 
or 

233.5    :    80.07    :  :    wt-  BaSO4  :    wt.   SO3 

weight  BaSO4  X  80.07  •   u      cr^ 

or          2 * •      —     weight    SO-, 

233-5 

and  to  get  percentage  we  have  : 

weight  taken  for  analysis  (magnesium  sulphate)  :  100  :  : 
weight  SO3  :  x 

weight  SOS  x   100  c  cr^ 

or  =  percentage  of  SOS.  u) 

weight  taken 

In  the  same  way  this  precipitate  of  BaSO4  can  be 
used  to  determine  the  percentage  of  barium  or  barium 
oxide,  here  we  have  for  barium  :  molecular  weight  of 


(*)  When  the  magnesium  sulphate  is  dissolved  in  water  it  is  partly  dissociated 
into  Mg  and  SO4  ions  which  react  with  the  Ba  and  Cl  ions,  so  the  statement  "mole- 
cule of  SO3  present  in  solution  "  is  not  literally  correct  but  should  be  SO4  ions 
corresponding  to  molecules  of  SO3.  This  does  not  however  affect  the  truth  of  the 
proportions  given. 


314  CALCULATIONS  FROM  EQUATIONS. 

BaSO4 :  atomic  weight  Ba : :  weight  BaSO4  :  weight  Ba, 

and  for  barium  oxide : 

mol.  wt.  BaSO4  :  mol.  wt.  BaO  :  :  wt.  BaSO4  :  wt.  BaO. 

or  233-5  :  i 53-43  :  :  wt.  BaSO4  :  x. 

and £ —  X  100=  Percentage  BaO. 

wt.  taken  for  analysis 

If  in  this  same  precipitate  the  sulphur  is  desired  (for 
example  in  coal  analysis)  we  have 

233-5  :  32-07  :  :  wt.  BaSO4  :  x  (wt.  S.) 

The  calculation  for  lead  sulphate  will  be  similar, 
though  the  proportion  will  be  different  on  account  of  the 
higher  atomic  weight  of  lead.  To  calculate  the  weight  of 
lead  corresponding  to  one  gram  of  PbSO4  we  have : 

302.99  (Mol.  wt.  PbSO4)  :  206.92  (At.  wt.  Pb)  :  :  i  :  x. 
x  =  0.6829. 

Suppose  we  precipitate  MgNH4PO4.  6H2O,  filter, 
wash,  ignite,  and  weigh.  On  ignition  this  is  converted 
into  magnesium  pyrophosphate,  Mg2P2O7,  which  can  be 
used  either  for  the  determination  of  magnesium  oxide 
or  phosphoric  anhydride.  We  have  the  reactions : 

MgCl,  +  NH4OH  +  Na2HPO4  =  MgNH4PO4  + 
2NaCl+H2O. 
and,  2MgNH4PO4  +  heat  =  Mg2P2O7+  2NH3+  H2O. 

For  every  molecule  of  the  ignited  precipitate  there 
must  have  been  present  in  the  original  solution  two 
atoms  of  magnesium,  two  atoms  of  phosphorus  or  one 
molecule  of  P2O5;  so  to  get  the  corresponding  amount 
of  MgO  we  have  : 

Mol.  wt.  Mg2P2O7 :  2  (Mol.  wt.  MgO) :  :  wt.  ppt.  :  x. 
or      222.6          :          80.58  :  :  wt.  ppt.  :  x. 

and   — ?— —  X  ioo  =  Percentage  of  MgO. 
wt.  taken 


CALCULATIONS  FROM   EQUATIONS.  35- 

Similarly  to  get  the  corresponding  weights  of  phosphorus- 
or  phosphoric  anhydride  we  have  : 

Mg-,P2O7  :  2?  :  :  wt.  ppt  :  x. 

222.6   :   62.02   :  :   wt.  ppt.   :  x. 

and  Mg,P,O7  :  P2O6        :  :  wt.  ppt.  :  x. 

222.6  :  142.02     :  :  wt.  ppt.    :  x. 

The  calculations  for  the  pyrophosphates  of  zinc  and 
of  manganese  are  similar,  using  the  atomic  weights  of 
these  metals  instead  of  that  of  magnesium. 

Zinc  is  often  weighed  as  ZnNH4PO4  :  in  this  case  for 
each  molecule  of  the  precipitate  there  will  be  but  one 
zinc  .  nd  the  proportion  will  be 

175.5  :  65-41  :  =  wt.  ppt.  :  x. 

All  direct  stoichiometrical  calculations  are  made  in 
this  way,  using  the  proper  atomic  weights  and  bear- 
ing in  mind  to  place  in  the  proportion  the  weight  of  the 
number  of  atoms  or  molecules  of  the  constituent  sought, 
which  the  precipitate  or  the  residue  weighed  contains. 

Indirect  Determinations. — In  the  material  weighed 
the  element  to  be  determined  may  not  be  present  at  all. 
For  example  :  Potassium  is  precipitated  quantita- 
tively as  K2PtCl6 :  this  may  be  weighed  as  such  or  may 
be  ignited  with  a  little  oxalic  acid,  the  soluble  potassium 
chloride  washed  out,  and  the  spongy  platinum  weighed. 
The  reactions  are 

2KC1+  H2PtCl6  =  K2PtCl6+  2HC1. 
To  calculate  the  weight  of  K2O  from  the  weight  of  the 
double  chloride  we  have 

Mol.  wt.  K3PtCl6  :  Mol.  wt.  K2O  :  :  wt.  ppt.  :  x. 
485.81  :          94.22  :  :  wt.  ppt.  :  x. 

If  platinum  is  weighed : — each  molecule  of  K2PtCl6 
has  one  atom  of  platinum  and  two  of  potassium,  so  for 


36  QUANTITY  OF  REAGENTS. 

each  atom  of  platinum  weighed  there  must  have  been 
two  atoms  of  potassium,  equivalent  to  a  molecule  of 
K2O  in  the  solution  to  cause  its  precipitation,  hence  we 
have 

At.  wt.  Pt  :  Mol.  wt.  K2O  :  :  wt.  residue  :  x 
or  194.89       :  94.22         :  :  wt.  residue  :  x 

and  in  either  case  -  —.  —  X  100  =  Percent  of  K2O. 
wt.  taken 

It  must  always  be  kept  in  mind  that,  when  there  is  a 
change  in  composition  during  ignition,  it  is  the  formula 
of  what  is  actually  weighed  which  gives  the  basis  for 
calculation.  For  example,  cobalt  is  often  separated 
from  nickel  by  precipitation  as  2Co  (NO2)3,  6KNO2  and 
after  filtering,  etc.,  is  converted  into  CoSO4  and  K2SO4 
by  sulphuric  acid.  The  properties  of  the  double  nitrite 
are  studied  for  the  purpose  of  separation  but  all  calcu- 
lations are  based  on  the  sulphates  weighed.  To  illus- 
trate :  One  gram  of  nickel  matte  was  taken  for  analysis 
and  the  mixed  sulphates  of  cobalt  and  potassium  weighed 
0.0896  gram.  What  ic  the  percentage  of  cobalt  ?  We 
have  the  reaction, 

2Co  (NO2)3,  6KNO2+  5H2SO4  =  2CoSO4 


so  for  every  2CoSO4  +  3K2SO4  there  must  be  2  Co  or 

832.87  :  117.86  :  :  0.0896  :  x.  x  =  0.0126  gram  or 
1.26%  Co. 

QUANTITY  OF   REAGENTS. 

In  analytical  as  well  as  technical  work  it  is  often  ne- 
cessary to  know  what  quantity  of  a  reagent  to  add  to 
cause  a  reaction. 

For  example.  How  much  4%  ammonium  oxalate 
is  necessary  to  precipitate  as  calcium  oxalate  the  lime  in 
one  gram  of  limestone  ?  We  have  the  reactions  ; 


QUANTITY  OF  REAGENTS.  37 

CaCO,  +  2HC1  =  CaCl.,  +  HoO  +  CO, 
CaCl,  +  (NH,)  C,O4  =  CaC,O4  +  2NH4C1. 

We  see  by  comparing  the  two  equations  that  one  mole- 
cule of  CaCO..j  requires  one  of  (NH4)C2O4.    So,  assuming; 
that  the  limestone  is  pure  calcium  carbonate,  we  have ; 
Mol.   wt.  CaCO3  :   Mol.   wt.  (NH4)  C2O4  :  :   wt.   lime- 
stone :  wt.  ammonium  oxalate  or 

100.09  :  126.16  ::  i  :  x.     x  =  1.2404  grams, 

but  a  4%  (by  weight)  solution  is  to  be  used ;  to  find 
what  weight  of  solution  contains  1.2404  grams  of  the  salt 
we  have 

1.2404  :  x  :  14:  100.     x  =  31.01  grams. 

It  is  not  to  be  inferred  that  this  is  necessarily  the 
best  amount  to  add  but  it  is  certain  that  less  will  be  in- 
sufficient for  complete  precipitation. 

In  the  precipitation  of  zinc  as  zinc  ammonium  phos- 
phate, it  is  considered  advisable  to  add  three  times  the 
amount  of  the  precipitant  actually  requ  red.  How  many 
grams  of  a  15%  solution  of  NH4NaHPO4  should  be 
added  to  a  solution  containing  0.25  gram  of  zinc?  The 
reaction  is : 

Zn(NO3)8+  NH4NaHPO4+  NH4OH.   = 

ZnNH4PO4  +  NH4NO3  +  NaNO3  +  H2O 

or  one  molecule  of  precipitant  for  one  atom  of  zinc : 
Hence  to  satisfy  the  equation  0.5245  gram  are  required 
for, 

I37-I5  :  65.41  : :  x  :  0.25.    x  =  0.5245. 
The  corresponding  weight  of  a  15%  solution  is  3.496 
grams,  for 

0.5245  :  x  :  :  15  :  100.    x  =  3.496. 
This  is  the  weight  of  a  1 5%  solution  necessary  to  satisfy 


210977 


38     TO  DETERMINE  WHICH  REAGENT  IS  MOST  ECONOMICAL. 

the  equation  but  we  are  to  add  three  times  this  amount 
or  10.488  grams. 

To  obtain  a  15%  solution  of  the  active  constituent: 
Microcosmic  salt  contains  four  molecules  of  water  of 
•crystallization,  so  to  make — say  100  grams  of  a  15% 
.-solution  we  must  first  find  what  weight  of  the  crystallized 
.salt  is  required  to  give  15  grams  of  the  active  con- 
stituent. 
Mol.  wt.  NaNH4HPO4  :  Mol.  wt.  NaNH4HPO4.  4H2O  :  : 

15  :  x     or  137.15  :  165.17  :  :  15  :  x.      x  =  18.064. 

So  we  must  weigh  out   18.064  grams  and  dissolve  it  in 
100-18.064  grams  or  81.936  grams  of  water. 

TO    DETERMINE    WHICH    REAGENT    IS    MOST    ECONOMICAL. 

Cases  often  arise  where  a  reaction  can  be  performed 
by  several  reagents,  so  that  it  is  of  importance  to  use 
the  most  economical,  not  necessarily  the  lowest  priced. 
For  example,  which  is  more  economical  as  an  oxidizing 
agent,  potassium  chlorate  at  8  cents  a  pound  or  sodium 
chlorate  at  10  cents? 

One  molecule  of  either  will  yield  three  atoms  of  oxy- 
gen, so  the  oxygen  in  a  pound  of  KC1O3  is  — - — .  Ib.  and 

122.56 

the  oxygen  in  a  pound  of  NaClO3  is— 1 —  Ib.     The  most 

106.5 

economical  is  the  one  which  gives  the  most  oxygen  for 
one  cent.     To  compare  these  we  have, 

KC1O3  — - —  -r-  8  or  0.0480  s  Ib.  for  one  cent 
122.56 

NaClO3      4       -r-  10  or  0.04507  Ib.  for  one  cent. 
106.5 

Which  shows  potassium  chlorate  to  be  more  economical 
at  the  prices  given. 


EXAMPLES. 


1.  What  is  the  weight  of  BaO  in  1.9327  grams  of 
BaCrO4  ?  Ans.     1.1694. 

2.  What  is  the  weight  of  CaO  in  2.9478  grams  or" 
CaSO4?  Ans..    1.2142. 

3.  What  weight  of  MnCO3  yields  on  ignition  1.450 
grams  of  Mn3O4?  Ans.     2.1847. 

4.  How  much  arsenic  is  there  in   5  grams  of  As2S3? 
ofAs2S5?   of  Mg2As2O7? 

Ans.      (a)  3.0463,  (b)  2.4167,  (c)  2.4151. 

5.  How  much  Agl  can  be  made  from  a  pound  of  pure 
silver?    How  much  AgBr?   How  much  AgCl? 

Ans.      2.1754  Ibs.  Agl,  1.7409  Ibs.  AgBr,  1.3285^5. 
AgCL 

6.  i.io  grams  of  stibnite  gave  on  analysis  0.5987 
gram    of  Sb2O4.     What  is  the  percentage  of  antimony  ? 

Ans.     43%- 

7.  How  many  pounds  of  water  will  it  take  to  slack 
loo  Ibs  of  quicklime  (CaO)?  Ans.      32.13  Ibs. 

8.  How  many  pounds  of  red  lead  (Pb;}O4)  can  be 
made  from  a  ton  of  litharge  (PbO)  ? 

Ans.      2048  Ibs. 

9.  How  many  grams  of  MnO2  and  of  H2SO4  are  re- 
quired to  give  10  grams  of  oxygen? 

Ans.      61.306  grams   H2SO4,  54.37  grams  MnO2. 

10.    One  gram  of  coal  gave  0.2634   gram  of  BaSO4 

iby  Eschka's  Method;    i  gram  of  MgO  and  0.5  gram  of 

(39) 


40  EXAMPLES. 

Na2CO3  were  used.  It  was  found  that  both  contained 
sulphur:  10  grains  of  MgO  and  5  grams  of  Na2CO3 
(together)  gave  0.1654  gram  of  BaSO4.  What  is  the 
percentage  of  sulphur  in  the  coal?  Ans.  3.39%. 

1 1.  How  many  pounds  of  20%  H2SO4will  it  take  to 
neutralize  a  ton  of  Na,CO:i  ?    of  N aH CO* ?  of  CaCO3  ? 

Ans.   (a)  9244  Ibs.,  (b)  5833.5  Ibs.,  (c)  9800  Ibs. 

12.  How  many  pounds  of  10%  HC1  will  it  take  to 
neutralize  a  ton  of  NaXO3?    ofNaHCO3?    ofCaCO3? 

Ans.  (a)  13744  Ibs.,  (b)  8673  Ibs.,  (c)  14571  Ibs. 

13.  Suppose  one  gram  of  silver  is  dissolved  in  nitric 
acid  and  to  it  is  added  0.25  gram   of  pure  dry  sodium 
chloride.      What  percentage  of  the  silver  remains  in  so- 
lution? Ans.      53-88%. 

14.  Suppose  0.25    gram    of  sodium  bromide   were 
added  to  a  solution  of  one  gram  of  silver.      What  per- 
centage of  silver  remains  in  solution  ? 

Ans.      73-81%. 

1 5.  A  dolomite  contains  98%  of  calcium   and  mag- 
nesium   carbonates,    2%  of  SiO2   and    10%  of  MgO. 
What  is  the  percentage  of  CO2?  Ans.     44.81%. 

1 6.  One  gram  of  ore  containing  nickel  and  cobalt 
gave  0.2750    gram    of  metallic  nickel  and   cobalt   and 
0.2103     gram    of  sulphates   of  potassium    and   cobalt 
(3K2SO4,2CoSO4)     What  is  the  percentage  of  nickel  ? 

Ans.     24.52%. 

17.  One  gram  of  a  rock  gave  on  analysis,  combined 
sodium  and  potassium   sulphates  o.  1 50  gram    and  plat- 
inum from  K2PtCl6  o. 1 127   gram.      What  are  the  per- 
centages of  K2O  and  Na.,O  ? 

Ans.     K,05.45%.     Na2O  2.15%. 

1 8.  In  an  iron  ore,  ferric  oxide,  alumina  and  phos- 
phoric   acid    were    precipitated    together,    ignited    and 


EXAMPLES.  41 

weighed  as  Fe2O3,  A12O3  and  FePO4.  The  weight 
from  three  grams  of  ore  was  2. 4750  grams:  the  iron 
was  found  to  be  50.5%  and  the  phosphorus  0.25%. 
What  is  the  percentage  of  A12O3?  Ans.  9.80%. 

19.  Which  is  more  economical  for  neutralizing  an 
alkali  60%  HNO3  at  6  cents  per  pound,  or  20%   HC1 
at  3  cents  per  pound?  Ans.     HC1. 

20.  Which  is  the  more   economical  oxidizing  agent 
KNO3  at  5  cents  a  pound  or  NaNO3  at  5^  cents   a 
pound?  Ans.     NaNO3. 


CHAPTER  V. 

CALCULATION  AND  USE  OF  FACTORS. 

IN  gravimetric  analysis  factor  means  the  number  by 
which  the  weight  of  any  precipitate  or  residue  is  multi- 
plied to  give  the  weight  of  the  constituent  desired, 
whether  the  constituent  is  actually  present  or  has  pre- 
viously been  combined  in  a  known  proportion. 

To  calculate  the  factor  for  SO3  in  BaSO4  we  have 
x  :  i  :  :  Mol.  wt  SO3  :  Mol.  wt.  BaSO4  or 
Mol.  wt.  SO3  80.07 


Mol.  wt.  BaS(4  o 

The  factor  is  the  weight  of  SO;5  in  one  gram  of 
BaSO4  or  one  hundreth  of  the  percent  of  SO3  in 
BaS04. 

To  calculate  the  factors  for  S  and  for  BaO,  we  have 

-¥^1   =  0.1373  and  I1M3  =  o.657i. 
233-50  233-50 

Factors  are  used  as  follows  :  — 
First,  To  simplify  calculations. 

Instead  of  making  the  proportion  — 

Mol.  wt.  BaSO4  :  Atomic  wt.  S  :  :  wt.  BaSO4  :  wt.  S. 
we  have  wt.  BaSO4  X  factor  (0.1373)  —  wt   S. 
So  that  if  we  have  a  table  of  factors,  the  calculation  re- 
quires only  one  multiplication,  and  this  can  be  changed 
into  an  addition  by  the  use  of  logarithms. 

These  tables  will  be  found  at  the  back  of  the  book. 

(42) 


CALCULATION  AND  USE  OF  FACTORS.  43 

Second,  To  avoid  or  to  greatly  reduce  calculation  by 
weighing  out  a  multiple  of  the  factor  for  analysis. 

If  the  weight  taken  in  grams  equals  the  factor  each 
milligram  of  precipitate  or  residue  equals  o.  i%  without 
any  calculation.  When  ten  times  the  factor  is  taken 
each  milligram  equals  0.01%.  When  one  hundred 
times  the  factor  is  used  each  milligram  equals  0.001% 
and  so  on. 

This  can  be  proved  as  follows : — 

Suppose  we  are  to  determine  SO3  in  crystallized 
magnesium  sulphate,  MgSO4,7H2O.  Weigh  out  for 
analysis  3.4291  grams  (ten  times  the  factor),  dissolve, 
and  then  precipitate  as  BaSO4,  in  the  usual  way ;  the 
precipitate  weighed  1.626  grams,  then  we  have 

1.626  x  0.3429  x  IOQ  =  Percentage  of  SQ3 

3-429 

but  0.3429  goes  into  3.429  ten  times  and  10  into  100  ten 
times,  so  the  result  in  percentage  is  ten  times  the  weight  in 
grams  or  16.26%,  or  the  precipitate  contained  1626  milli- 
grams, each  of  which  is  therefore  equivalent  to  0.01%. 

Again  the  factor  for  S  in  BaSO4  is  0.13734.  If  we 
weigh  out  13.734  grams  of  pig  iron  for  analysis,  each 
milligram  of  BaSO4  will  be  equivalent  to  0.001%  S. 
For  suppose  the  BaSO4  obtained,  weighed  0.0270  gram  ; 
we  would  have 

0.027x0.13734  x  I00  =  percent.  of  S  or  0.027%. 

13-734 

So  27  milligrams  =  27  thousandths  percent  or  each 
milligram  =  o.ooi  percent.  But  this  is  too  large  an 
amount  to  take  unless  the  sulphur  is  extremely  low,  so 
in  practice  some  portion  of  it  is  used  as  ^,  ^  or  0.4 ; 
then  the  weight  of  BaSO4,  expressed  in  grams,  times 
2,  3,  or  2.5  is  the  percentage  of  sulphur  without  any 


44 


CALCULATION  AND  USE  OF  FACTORS. 


further  calculation.  For  example,  if  for  the  determin- 
ation of  sulphur,  5.4936  grams  is  taken,  the  weight  of 
BaSO4  in  grams  divided  by  four  and  multiplied  by  ten 
gives  the  percentage  of  sulphur,  for  the  weight  5.4936 
is  forty  times  the  factor  for  S  in  BaSO4.  Suppose  the 
actual  weight  of  BaSO4  to  be  0.200  gram,  then 

020oxai373_4  =  aoo5  wt.  S  in  one  gram, 


and  this  times  100  gives  percentage  or  0.5%  ; 

which  result  is  a2O°X  10  =  0.5%. 

4 

Inasmuch  as  these  relations  are  confusing  at  first, 
especially  when  the  weights  of  precipitates  are  sometimes 
given  in  grams,  and  sometimes  in  milligrams,  the  follow- 
ing table  may  prove  useful. 


Weight  taken 
in  Grams. 

Weight  Precipitate 
in  Grams. 

Weight  Precipitate  in  Milligrams. 

Factor 

Times  100  =  Percent 

Divided  by  10  =  Percent 
EachMg.  =  0.1% 

Factor  x  5. 

Times  20  =  Percent 

Divided  by  50  =  Percent 
Each  Mg.  =  0.02% 

Factor  x  10 

Times  10  =  Percent 

Divided  by  100  =  Percent 
Each  Mg.  =  0.01% 

Factor  x  50 

Times  2  =  Percent 

Divided  by  500  =  Percent 
Each  Mg  =  0.002% 

Factor  X  100 

Percent  Direct 

Divided  by  1000  =  Percent 
Each  Mg  =  0.001% 

The  use  of  factors  to  save  time  should  not  be  pushed 
too  far,  as  the  time  taken  in  obtaining  an  exact  quantity  of 
some  crystallized  salts  or  of  drillings  may  be  greater  than 


CALCULATION  AND  USE  OF  FACTORS.          45 

that  saved  in  calculation.  The  examples  given  have 
that  objection,  which  however  would  not  apply  to  an  iron 
ore  or  coal  in  which  sulphur  was  to  be  determined,  as 
the  sample  would  be  in  a  finely  pulverized  condition.  It 
takes  but  little  longer  to  weigh  out,  say  exactly  1.3734 
grams  than  exactly  one  gram,  for  the  time  is  consumed 
in  getting  the  exact  weight,  not  in  putting  the  weights 
on  the  pan :  and  when  an  analysis  is  repeatedly  made, 
even  this  time  can  be  saved  by  using  a  special  weight, 
the  most  convenient  multiple  of  the  factor. 

This  system  can  be  very  easily  applied  also  to  volu- 
metric work,  here  the  amount  to  be  weighed  out  will  be 
a  multiple  of  the  standard  (liter)  of  the  solution  used. 
Suppose  we  have  a  solution  of  KMnO4  i  c.c.  =  0.0056 
gram  of  iron.  If  we  take  for  analysis  0.56  gram  of 
iron  ore  and  55.1  c.c.jof  KMnO4  are  used,  the  ore  con- 
tains 55.1%  ofTron  for 

55.1   x  0.0056  Xl  %. 

0.50 

So  that  when  we  take  for  the  test  a  weight  equal  to 
hundred  times  the  standard  of  the  solution,  each  cubic 
centimeter  equals  one  percent  and  the  burette  reading 
gives  percentage  direct. 

One  hundred  times  the  standard  may  not  be  the  most 
convenient  quantity  to  take,  but  this  can  be  readily  be 
adjusted  to  the  particular  titration,  bearing  in  mind  that, 
when 

50  times  the  standard  is  taken  i  c.c.  =  2%. 
100      "       "          "  "  i  c.c.  =  i%. 

200      "        "       ,  "  "  i  c.c.  =  0.5%. 

and  so  on. 

The  plan  of  adjusting  the  weight  taken  to  the 
strength  of  the  solution  presents  many  advantages  over 


46          CALCULATION  AND  USE  OF  FACTORS. 

the  system  of  adjusting  the  solution  to  a  certain  weight 
for  analysis,  when  a  portion  is  used  for  a  single  deter- 
mination. This  is  particularly  the  case  when  the  standard 
solution  changes  in  strength  and  has  to  be  frequently  re- 
standardized.  If  we  always  use  one  burette,  which  is 
uniformly  calibrated,  for  the  particular  solution  and 
weigh  out  a  multiple  of  the  standard  for  analysis,  that 
burette  will  read  percentage :  and  we  do  not  have  to 
make  up  the  solution  to  a  certain  arbitrary  strength, 
which  necessitates  accurate  measuring  apparatus,  flasks, 
etc.,  which  agree  with  the  burette.  Then  if  on  standing 
the  solution  evaporates  or  undergoes  a  partial  decom- 
position, it  must  be  readjusted  to  read  percentage  when 
an  arbitrary  weight  is  taken,  while  by  weighing  out  a 
multiple  of  the  new  standard  much  trouble  is  saved  with 
no  sacrifice  of  accuracy.  It  is  not  to  be  inferred  how- 
ever that  this  system  is  advantageous  in  all  cases,  but 
only  for  separate  determinations  such  as  iron,  copper, 
zinc,  etc. 

The  most  important  application  of  the  use  of  factors, 
is  the  assay  ton  system  devised  by  Prof.  Chandler.  In 
assaying  ores  of  gold  and  silver,  the  precious  metals  are 
always  reported  in  ounces  Troy  per  ton  of  2000  pounds 
avoirdupois.  If  a  suitable  number  of  grams,  twenty-five 
for  instance,  is  taken  and  the  gold  and  the  silver  weighed 
in  milligrams,  a  long  calculation  is  required  to  convert 
the  results  to  ounces  per  ton.  If  however  an  assay  ton 
29.1666  grams,  is  used,  the  result  in  gold  or  silver  as 
weighed  in  milligrams  is  ounces  to  the  ton  without  any 
calculation. 

The  assay  ton  is  derived  as  follows : — 
One  pound  avoirdupois  contains  7000  grains. 
One  ton,  2000  Ibs.,  contains  14,000,000  grains. 
One  ounce  Troy  contains  480  grains. 


CALCULATION  AND  USE  OF  FACTORS.         47 

Therefore  14,000,000  -=-  480  or  29,166.6  is  the  number 
of  Troy  ounces  in  2000  Ibs.  avoirdupois ;  so  if  we  take 
this  number  of  milligrams  for  assay  (29,166.6  -f-)»  each 
milligram  of  gold  or  of  silver  found  is  an  ounce  Troy  to 
the  ton  in  the  ore  :  for 

i  :  29,166.6  +  :  :  480  :  14,000,000. 

Therefore  the  assay  ton  is  29,166.6  +  milligrams  or 
29.1666  +  grams. 


EXAMPLES. 


1.  Calculate  the  factors  for  Na2O  in  Na2SO4  and  in 
NaCl.  Ans.     0.4368  and  0.5308. 

2.  Calculate  the  factors  for  Zn   in  ZnO,  in  ZnNH4 
PO4  and  in  Zn2P2O7. 

Ans.     0.80346,  0.3664  and  0.4291. 

3.  Calculate  the  factors  for  As  in  As2S3,  in  Mg2As2O7 
and  Ag3AsO4.  Ans.     0.6092,  0.4830  and  0.1621. 

4.  Calculate  the  factors  for  N,  NH3  and  NH4C1,  for 
weight  of  platinum  from  ignition  of  (NH4)2PtCl6. 

Ans.     0.1441,  0.1751  and  0.54925. 

5.  What  are  the  factors  for  conversion  of  Cr  into 
Cr2O3,  into  PbCrO4  and  into  BaCrO4  ? 

Ans.     1.4603,  6.1960  and  4.8632. 

6.  (a)  What  is  the  factor  to  conyert  weight  of  BaSO4 
to  corresponding  weight  of  H2SO4? 

(b)  What  is  the  factor   to    convert   percentage   of 
K2O  to  percentage  of  K2SO4? 

Ans.     (a)  0.42007.     (b)  1.8498. 

7.  What   are  the   factors   for  K2O  and  for  KC1    in 
platinum  from  K2PtCle? 

Ans.     0.48345  and  0.76515. 

8.  What  are  the  factors  for  the  iron  in  Fe2O3  and  in 
Fe3O4?  Ans.     0.7001  and  0.7242. 

9.  One    gram    of  an    antimony  ore   was   taken   for 
analysis  and  the  SboO4  resulting  weighed  0.3478  gram: 

(48) 


270 


EXAMPLES.  49 

by  mistake   the  factor  for  Sb2S3  was  used.     What  was 
the  error  in  percentage  of  antimony  ? 

Ans.     2.626%. 

10.  An  alloy  contains  10%  of  antimony.  One  gram 
of  this  alloy  gives  0.2895  gram  of  combined  oxides  of 
antimony  and  tin  (Sb2O4  and  SnO2).  What  would  be 
the  error  in  percentage  of  tin  if  only  the  factor  for  SboO4 
were  used  in  the  calculation  ?  Ans.  0.032%. 

1 1.  What  weights  of  spiegel  must  be  taken  for  ana- 
lysis so  that  each  milligram  of  Mn2P2O7  shall  equal  re- 
spectfully 0.2%,  0.1%  and  0.05%? 

Ans.     0.1936,  0.3872  and  0.7744. 

12.  What  weights  of  sphalerite  must  be  used  so  that 
each    milligram    shall   equal    o.  i%   when    the   zinc    is 
weighed  as  ZnO,  as  ZnNH4PO4  and  as  Zn2P2O7? 

Ans.     0.8035,  0.3664  and  0.4291. 

13.  What  weight  of  steel  must  be  taken  so  that,  when 
the  phosphorus   is   weighed  as  Mg2P2O7,  the  weight  in 
grams  shall  be  percentage  direct?       Ans.     27.8681. 

14.  What  weight  of  pig  iron  must  be  taken  so  that  the 
weight  of  BaSO4  in  grams  shall  be  percentage  of  sul- 
phur when  multiplied  by  three.  Ans.     4.5780. 

15.  What  weights  of  lead  ore  must  be  used  so  that 
each  milligram  of  PbSO4  shall  equal  0.05%,  o.  i%  and 
0.2%?  Ans.      1.3658,  0.6829  and  0.3415. 

1 6.  What  weights  of  limestone  must  be  taken  so  that 
each  milligram  of  CaO  equals  o.  i%  of  CaO  ?    So  that 
each  milligram  of  CaSO4  equals  0.1%  of  CaO  ? 

Ans.     i  and  0.41 19. 

1 7.  Given  a  solution  of   K2Cr2O7,  i   c.c.  =  0.0054 
gram   of  iron.      How   much  ore  must  be  taken  so  that 
each  c.c.   shall   read   0.1%,  0.2%,  0.3%,  0.5%,  0.7% 
and  i.%?          Ans.     5.4,  2.7,  1.8,  1.08,  0.7714,  0.54. 


'50  EXAMPLES. 

1 8.  Given  a  solution  of  K4Fe(CN)6,  i  c.c.  =  0.00896 
gram  of  zinc.      How  much  ore  must  be  taken  so  that 
each  c.c.  shall  equal  2%,  i%,  0.5%,  o.  i%? 

0.4480,  0.8960,  1.792,  8.960. 

19.  Calculate  the  ounces  Troy  per  ton  avoirdupois 
in  the  following — 

(a)  One  tenth  assay  ton  of  copper  matte  gave  gold 
1.22  mgs.,  silver  30.94  mgs. 

(b)  Four  assay  tons  of  siliceous  ore  gave  gold  o.  1 2 
mg.,  silver  0.52  mg. 

(c)  One  sixth  assay  ton  of  a  zinc  blende  gave   gold 
0.02  mg.,  silver  37.72  mgs. 

Ans.\ 

,     \  12.2  ozs.  Au.  xyx  (  0.03  oz.Au.   ,  ^(0.12  oz.  Au. 
}  309.4028.  AgA  '  \  o.  13  oz.  Ag.  ^  '  \  226. 32  ozs. Ag. 

20.  Calculate  the  number  of  grams  in  an  assay  ton 
for  a  long  ton  (2240  pounds). 

Ans.      32.6666  grams. 


CHAPTER   VI. 

CALCULATIONS   OF   VOLUMETRIC    ANALYSIS. 
NORMAL  SOLUTIONS. 

A  normal  solution  is  one  which  contains  the  hydro- 
gen equivalent  of  the  active  constituent  in  grams  per 
liter.  That  is,  the  amount  in  a  liter  which  brings  into  re- 
action 1.008  grains  of  hydrogen,  8  grams  of  oxygen  or 
their  equivalents,  whether  the  reaction  ^be  one  of  oxida- 
tion, reduction,  precipitation  or  saturation. 

This  definition  is  used  by  Mohr  and  by  Sutton  and  is 
also  in  very  general  use  in  America.  In  the  examples 
given  in  this  book  this  system  is  used.  There  are  how- 
ever other  definitions  of  normal  solutions,  which  have 
caused  considerable  confusion.  They  are : 

First.  The  molecular  weight  in  grams  per  liter.  This 
definition  possesses  some  advantages,  for  example  ac- 
cording to  Mohr's  definition  we  can  have  two  normal 
solutions  of  KMnO4  of  different  strengths  depending  on 
whether  it  is  used  in  a  neutral  or  in  an  acid  solution ; 
while  according  to  the  "  molecular  weight  in  grams  "  this 
can  not  happen,  on  the  other  hand  Mohr's  definition  has 
the  great  advantage  that  all  normal  solutions  are  equiva- 
lent cubic  centimeter  for  cubic  centimeter.  Solutions 
containing  the  molecular  weight  in  grams  per  liter  are 
now  much  used  in  physical  chemistry  under  the  name 
gram-molecule. 

Second.  In  the  Gay-Lussac  method  for  the  determ- 
ination of  silver,  used  in  mints  and  assay  offices,  normal 

(51) 


52  NORMAL  SOLUTIONS. 

salt  solution  is  the  name  given  to  a  solution  of  such 
strength  that  one  hundred  cubic  centimeters  will  exactly 
precipitate  one  gram  of  silver.  This  is  a  special  use  of 
the  word  normal  and  should  not  be  connected  in  any 
way  with  the  equivalent  system  of  volumetric  analysis. 
Under  the  system  we  have  adopted,  it  is  absolutely 
essential  to  have  a  correct  understanding  of  "the  hydro- 
gen equivalent  in  grams."  Suppose  we  wish  to  make  a 
normal  solution  of  sodium  hydroxide ;  we  have  the  re- 
action— 

NaOH+HCl=NaCl  +  H2O. 

What  is  the  amount  necessary  to  bring  into  reaction 
one  equivalent  of  hydrogen  ?  Evidently  the  molecular 
weight  in  grams  of  NaOH  (40.06)  for  this  is  the  amount 
necessary  to  react  with  36.46  grams  of  HC1  and  so 
bring  into  reaction  1.008  grams  of  hydrogen. 

If  we  wish  a  normal  solution  of  Na2CO3  we  have 
Na2CO3  +  2HC1  =  2NaCl  +  H2O  +  CO2.  It  is  evident 
that  one  molecule  of  Na2CO3  brings  into  reaction  two 
atoms  of  hydrogen,  hence  a  normal  solution  of  Na2CO3 
will  be  one  half  of  the  molecular  weight  or  53.05  grams 
per  liter. 

Similarly  a  normal  solution  of  hydrochloric  acid  will 
contain  the  molecular  weight,  36.46  grams  per  liter,  of 
sulphuric  acid  one  half  the  molecular  weight,  49.043 
grams,  and  of  orthophosphoric  acid  one  third  the  mole- 
cular weight  in  grams  or  32.68  grams. 

One  cubic  centimeter  of  any  of  these  acid  solutions 
will  exactly  saturate  one  cubic  centimeter  of  any  of  the 
alkaline  solutions. 

In  reactions  of  precipitation  the  same  rule  applies 
we  have, — 

AgN03  +  HC1  =  AgCl  +  HN03  or 
AgNO3  -r-NaCl  =  AgCl  +  NaNOs 


NORMAL  SOLUTIONS.  53 

A  normal  solution  of  silver  nitrate  contains  the  molecu- 
lar weight  in  grams,  169.96,  per  liter  :  for  according  to 
the  first  reaction  1.008  grams  of  hydrogen  would  be 
actually  brought  into  reaction  by  a  liter  of  this  solution 
and  in  the  second,  the  weight  of  sodium  equivalent  to 
1.008  grams  of  hydrogen.  Similarly  with,  NH4SCN-f- 
AgNO3=  AgSCN+NH4NO3,  a  normal  solution  will 
contain  the  molecular  weight  of  ammonium  thiocyanate 
in  grams  per  liter  or  76. 19  grams.  To  precipitate  a  salt 
of  a  diatomic  element  such  as  BaCrO4,  we  have  a  normal 
solution  equal  to  one  half  the  molecular  weight  in  grams 
per  liter;  for  K2CrO4,  97.18  grams;  for  Na-jCrO^  81.12 
grams. 

For  convenience  we  designate  a  normal  solution  by 
N,  tenth  normal  by  N/10,  one  hundreth  normal  by  N/100 
and  .so  on. 

In  the  original  definition  of  normal  solution  appears 
"  equivalent  to  one  gram  of  hydrogen."  This  can  be 
retained  with  perfect  accuracy  if  the  atomic  weights  used 
are  calculated  on  the  basis  of  hydrogen  as  unity  ;  but 
as  the  ratio  of  hydrogen  to  oxygen  has  been  found  to 
be  i  :  15.88  and  not  i  :  16  we  must  adopt  either  H  =  i 
or  O  =  1 6  as  a  basis.  O  =  16  is  most  generally  used  at 
present  so  to  make  our  definition  of  normal  comparable 
with  the  present  atomic  weights,  instead  of  equivalent 
to  one  gram  of  hydrogen,  we  must  substitute,  equivalent 
to  i. 008  grams  of  hydrogen  or  better  equivalent  to  8 
grams  of  oxygen.  The  advantage  of  the  latter  definition 
will  appear  in  the  following  explanations  of  normal  solu- 
tions for  oxidation  and  reduction. 

When  ferrous  iron  is  oxidized  by  potassium  dichro- 
mate  we  have  the  reaction  K2Cr2Or-f3Fe2Cl4  +  I4.HC1 
=  3Fe2Cl6-f  2KC1+  Cr,Cl6  +  i4H,O  :  the  essential  oxi- 
dation in  this  reaction  can  be  expressed  as  follows — 


54  NORMAL   SOLUTIONS. 

6FeO  -|-  3O  =  3Fe2O3.  Three  atoms  of  oxygen  from 
the  K2Cr2O7  oxidize  six  atoms  of  ferrous  iron  as  chloride 
to  the  ferric  condition,  or  in  other  words  an  iron  requires 
one  half  an  oxygen,  equivalent  to  one  hydrogen  :  so  for 
a  normal  solution,  the  molecular  weight  of  K2Cr2O7  in 
grams  must  be  divided  by  six,  since  each  molecule  oxi- 
dizes six  atoms  of  iron  each  equivalent  to  one  atom  of 
hydrogen  or  one  half  an  atom  of  oxygen,  or  expressed 
in  the  form  of  a  proportion  we  have — 

Mol.   wt.  K2Cr2O7:  x::  3  x  16:  8.     x  =  49.083  grams. 

With  KMnO4  we  have  2KMnO4  -f  sFe2  (SO4)2  + 
8H2SO4  =  5Fe2  (SO4)3  +  K2SO4  +  2MnSO4  +  8  H2O. 

The  essential  oxidation  is  Mn2O7-f-  loFeO  =  5Fe2O3 
+  2MnO  or  2KMnO4  give  5  oxygen.  To  calculate 
the  oxygen  equivalent  we  have,  2  (Mol.Wt.  KMnO4) 
=  316.2  grams,  gives  oxygen,  5X16=80  grams, 
hence  316.2  :  x  :  :  80  :  8  or  a  normal  solution  of  KMnO4 
contains  31.62  grams  per  liter  when  used  in  an  acid  solu- 
tion. 

In  the  Volhard  method  for  manganese  we  have  in  a 
neutral  solution  the  following  reaction 
2KMnO4+3MnSO4+2ZnO  = 

5MnO2+  K2SO44-ZnSO4 

the  essential  oxidation  isMn2O7  +  3MnO  =  5MnO2  or 
two  molecules  of  KMnO4  give  up  under  these  conditions 
three  atoms  of  oxygen,  so  to  get  a  normal  solution  we 
have  316.2  :  x  :  :  48  :  8  or 

x  =  Mol.Wt  -v-6  =52.7  grams  per  liter. 

This  is  one  of  the  very  few  cases  where  the  molecu- 
lar weight  definition  of  normal  seems  preferable,  as  here 
we  have  two  normal  solutions  of  different  strengths  de- 
pending on  the  reaction.  As  permanganate  is  usually 
used  in  an  acid  solution,  unless  expressly  stated,  normal 
means  31.62  grams  per  liter. 


NORMAL   SOLUTIONS.  55 

Normal  reducing  solutions  are  not  much  employed, 
but  the  following  will  serve  as  an  illustration.  We  have 
the  reaction 

SnCl2  +  Fe2Cl6  =  SnCl4  +  Fe2Cl4 : 

one  molecule  of  SnCl2  removes  two  atoms  of  chlorine  or 
what  is  the  same  thing  reduces  two  atoms  of  iron  to  the 
ferrous  condition.  Now  two  chlorine  correspond  to  one 
oxygen  ;  hence  one  half  the  molecular  weight  in  grams 
of  stannous  chloride  will  give  a  normal  solution  or 

189.95  :  x  :  :  16  :  8.    x  =94.975  grams. 

To  ascertain  the  strength  of  any  normal  solution  the 
reaction  in  which  it  is  to  be  used  must  always  be  consid- 
ered and  in  this  way  all  confusion  will  be  avoided. 

An  advantage  of  the  "  equivalent "  definition  of  nor- 
mal, which  deserves  attention,  is  that  when  the  formula 
of  a  compound  is  altered  the  equivalent  is  not ;  so  that 
whether  you  consider  potassium  permanganate  as  K2Mn2 
O8  or  KMnO4,  arsenious  oxide  as  As2O3  or  As4O6  or  fer- 
ric chloride  as  Fe2Cl6  or  FeCl3,  the  number  of  grams  per 
liter  of  a  normal  solution  will  remain  the  same.  This  is 
made  clear  by  the  following  reactions. 
KoMnsO8  +  ioFeSO4  +  8H2SO4  = 

5Fes(SO4)3+  K2SO4+  2MnSO4+ 8H2O 
Now  written 
2KMnO4  +  5Fe2  (SO4)2  +  8H2SO4  = 

5Fe2(SO4)3  +  K2SO4+  2MnSO4-j-8H2O 

In  either  case  the  same  weight  of  potassium  perman- 
ganate oxidizes  the  same  weight  of  iron  and  normal  solu- 
tions will  be  identical  for  N  K2Mn2O8  will  be  one  tenth 
the  molecular  weight,  31.62  grams,  and  N  KMnO4  one 
fifth,  31.62  grams.  The  same  applies  to  ferrous  sulphate. 

In  using  normal  solutions  the  following  rules  are  im- 
portant. Weigh  out  one-tenth  the  equivalent  in  grams 


56  NORMAL    SOLUTIONS. 

of  the  substance  to  be  determined  and  the  burette  reading 
gives  the  percentage  directly.  With  N/10  solutions  weigh 
out  one  hundreth  of  the  equivalent  in  grams  and  the 
burette  shows  the  percentage  directly.  To  make  this 
clear,  suppose  we  have  a  normal  sulphuric  acid  solution 
and  we  wish  to  titrate  a  sample  of  caustic  potash  and 
to  express  the  results  in  percentage  of  K2O.  N  H2SO4 

contains  ^  '°^    grams  per  liter   or    i.   c.c.   =  0.049045 

gram  of  H2SO4.     The  equivalent  of  K,O  is^li^-2    or 

47.1 1 ;  so  N  H2SO4  is  equivalent  to  0.0471 1  gram  K2O 
per  c.c.  One-tenth  the  equivalent  in  grams,  4.711, 
is  weighed  out,  dissolved  and  titrated,  requiring 
78.2  c.c.  of  N  H2SO4.  Obtain  the  percentage  by  go- 
ing through  the  ordinary  calculation 

78.2X0.047"  x  I00  =  ;8.2%. 
4.711 

It  is  seen  that  this  is  only  another  way  of  expressing  the 
fact  that  when  one  hundred  times  the  standard  is  taken 
for  analysis,  the  burette  gives  percentage  direct.  ^ 

Besides  normal  solutions  we  have  standard  solutions, 
that  is  solutions  of  known  strength.  Normal  solutions 
are  always  standard  but  standard  solutions  are  not 
always  normal.  1  he  salt  solution  used  in  the  mints  is 
standard,  but  only  normal  in  a  particular  sense  of  the 
word.  Most  of  the  solutions  used  in  volumetric  analy- 
sis are  standard  and  only  approximately  normal,  tenth 
normal  or  one  hundreth  normal.  The  question  of  the 
strength  of  solutions  and  whether  systematic  or  not  is 
largely  a  matter  of  individual  preference,  limited  by  the 

(*)  A  table  of  the  more  common  N  and  N/10  "solutions  will  be  found  at  the 
back  of  the  book. 


ALKALIMETRY    AND    ACIDIMETRY  57 

solubility  of  the  reagent  and  the  requirements  of  the  par- 
ticular analysis  to  be  performed. 

In  the  calculations  of  volumetric  as  in  gravimetric 
analysis  the  basis  of  calculation  is  an  equation.  Unfort- 
unately there  are  some  cases  in  which  the  reaction  can 
not  be  expressed  by  a  simple  equation  and  in  these  cases 
standard  not  normal  solutions  must  be  used. 

ALKALIMETRY    AND    ACIDIMETRY. 

Many  of  these  calculations  require  no  special  men- 
tion here  as  they  can  readily  be  performed  after  the 
explanations  already  given.  Only  a  few  special  cases 
are  given  in  detail. 

In  order  not  to  complicate  these  calculations  by  the 
use  of  different  indicators,  phenolphtalein  is  used  through- 
out. This  indicator  acts  as  follows : 

Red  with  fixed  alkalies,  colorless  with  acids,  includ- 
ing carbonic.  So  when  acid  is  run  into  carbonated 
alkali  the  color  is  discharged  when  the  Na2CO3  is  con- 
verted into  NaHCO.3,  but  on  boiling  the  color  returns 
due  to  the  formation  of  more  Na2CO3  from  the  NaHCO3, 
so  to  get  total  Na->CO3,  acid  must  be  run  in  till  the  solu- 
tion remains  colorless  ,on  boiling.  The  reactions  are 

NaXO3+  HC1  =  NaHCO3+  NaCl  and  (boiling) 
NaHCO3  +  HC1  =  NaCl  +  CO2  +  H2O. 

i.   Caustic  and  Carbonated  Alkali. 

1.5  grams  of  impure  caustic  potash  are  dissolved  in 
water,  diluted,  an  excess  of  barium  chloride  added 
and  diluted  to  400  c.c.  One  hundred  cubic  centimeters 
are  filtered  off  and  titrated  with  N  HC1,  5.4  c.c.  required. 
Another  1.5  grams  of  the  sample  dissolved  and  titrated 
directly  with  N  HC1  boiling,  for  total  alkalinity,  re- 
quired 24.6  c.c.  What  are  the  weights  of  KOH  and 


58  ALKALIMETRY   AND   ACIDIMETRY. 

ofK2CO3?  The  first  titration  gives  the  caustic  alkali 
m  — —  grams,  for  the  barium  chloride  precipitates  the 

carbonate  as  barium  carbonate  and  with  the  caustic  gives 
equivalent  amount  of  Ba  (OH)2,  which  is  determined  by 
the  N  HC1.  The  reactions  are 

Na2CO3  +  BaCl2  =  BaCO3  +  2NaCl  and 
2NaOH  +  BaCl2  =  Ba(OH),  +  2NaCl 
The  total  required          24.6  c.c. 
The  caustic  required      21.6  c.c. 
The  carbonate  required  3.0  c.  c 
N  HC1  =  0.05612  gram  KOH  per  c.c. 

=  0.0691 1  gram  K2CO3  per  c.c. 

hence  the  weights  are  KOH,  1.2122  grams  and  KsCOa 
0.2073  gram. 

2.   Carbonate  and  Bicarbonate. 

Four  grams  of  commercial  sodium  bicarbonate  are 
titrated  in  a  cold  dilute  solution  with  N/10  H2SO4  till  the 
red  color  disappears :  used  6.4  c.c. 

To  one  gram  of  the  sample  dissolved  in  water,  20  c.c. 
of  N  HoSO4  are  added  and  the  solution  boiled  till  all 
the  carbonic  acid  is  expelled,  then  titrated  back  with 
N  KOH  :  requiring  8.1  c.c. 

What  are  the  percentages  of  NaHCO3  and  of 
NaX03? 

The  first  titration  shows  one  half  of  the  Na2CO3,  so 
for  one  gram  the  Na2CO3  required 

2  x  6'4  =3.2  c.c.  of  %  H2SO4  or  0.32  c.c.  N  H2SO4. 

4 

The  total  Na2O  required  20-8.1  =  1 1.9  c.c.  N  H2SO4 
so  the  N  H2SO4  used  on  the  Na2O  present  as  NaHCO3 
was  1 1.9  —  0.32  =  1 1.58  c.c. 


ALKALIMETRY   AND   ACIDIMETRY.  59 

The  weights  corresponding  are 

NaHCO3  11.58  X  0.08406  =  0.9734  gram 
Na2CO3    0.32  X  0.053        =  0.0169  gram 
or  97-34%  NaHCO3  and  1.69%  Na2CO3. 

3.   Phosphorus  from  Ammonium  Phospho-Molybdate. 

The  phosphorus  is  precipitated  as(NH4)3PO4,  i2MoO3, 
under  carefully  regulated  conditions,  this  is  dissolved 
in  a  standard  alkali  solution  according  to  the  reaction  ; 


2(NH4)2HPO4+(NH4)2Mo4+23Na2MoO4+23CO., 

In  practice  an  excess  of  the  standard  Na2CO3  solu- 
tion is  added  then  the  excess  titrated  using  phenol- 
phtalein  as  an  indicator.  The  reason  that  two  molecules 
of  the  precipitate  require  23  and  not  24  molecules  of  Na2 
CO3  is  that  the  indicator  changes  color  when  two  of  the. 
three  replaceable  hydrogens  of  phosphoric  acid  have 
been  substituted  by  alkali,  which  leaves  (from  two  mole- 
cules) two  NH4  groups  to  combine  with  a  molecule  of 
MoO3  thus  leaving  23  instead  of  24  molecules  of  MoO3 
to  be  neutralized  by  Na2CO3. 

To  calculate  the  grams  per  liter  of  Na2CO3  solution 
equivalent  to   one   milligram  of  P2O5  per  cubic   centi- 
meter, we  have:    i.c.c.  =  i  mg.      i  liter  =  i  gram. 
2(NH4)3PO4,  i2MoO3   contain    iP2O5  hence 

Mol.  wt.   P2O5  :    23(Mol.wt.Na2CO3)  :  :  i  :  x 
142.10         :  2440.3  :  :  i  :  x 

x=i7.  1  80  grams  per  liter  of  Na2CO3. 

4.   Sulphuric  and  Nitric  Acids  in  Mixed  Acid. 

The  analysis  of  the  waste  mixed  acid  from  nitrating 
is  sometimes  made  as  follows  :  50  grams  of  the  sample 
are  diluted  to  a  liter  and  10.  c.c.  of  this  mixture  are  with- 


60  PERMANGANATE   CALCULATIONS. 

drawn  by  a  pipette  which  in  this  case  delivered  an  amount 
containing  exactly  0.4976  gram  of  the  original  sample. 
The  sulphuric  acid  in  this  is  precipitated  as  BaSO4  and 
weighed  =  0.7093  gram. 

Another  portion  is  titrated  for  total  acidity  as  follows: 
Of  an  approximately  N  NaOH  solution  (20.2  c  c.  =  0.98 
gram  H2SO4)2o.2  c.c.  are  run  into  a  beaker  and  this  is 
the  neutralized  by  the  diluted  acid  solution.  (50  grams 
to  liter.)  Suppose  25.75  c-c-  are  required. 

To  calculate  the  percentage  of  H.3SO4 

o. 7093  x.  c 142007 x, oo  ='59.88% 

0.4976 

To  calculate  the  percentage  of  HNO3 
25.75  c.c.  =  20.2  c.c.  NaOH  =  0.98  gram  H2SO4 

First  calculate  the  total  acidity  to  HoSO4 : 
As  50  grams  are  contained  in   1000  c.c.    25.75  c.c. 
contain  1.2875  grams  for  50:    1000:  :  x  :  25.75     or  x  = 

5'/5_  1.2875  grams.     Hence  the  total  acidity  in  terms 
20 

of  H2SO4  is  °'98      X  ioo   or  76.12%  and  76.12%    - 
1.2875 

59.88%  or  16.24%  is  tne  acidity  due  to  nitric  acid  but 
expressed  in  terms  of  H2SO4.  As  63.048  parts  of  HNO:. 
are  equivalent  in  power  of  neutralizing  NaOH  to  49.043 
parts  of  HoSO4  ;  to  determine  the  percentage  of  HNO3 
we  have  :  16.24  :  x  ::  49-°43  :  63.048.  x  =  20.87%. 
So  the  composition  of  the  waste  acid  is 

H2S04   59-88%,  HN03  20.87%. 

PERMANGANATE    CALCULATIONS. 

Suppose  we  have  a  N/5  solution  of  KMnO4  ;  to  cal- 
culate its  strength  in  terms  of  oxalic  acid  we  have  the  re- 
actions. 


PERMANGANATE   CALCULATIONS.  61 

2KMnO4+  5Fe2(SO4)2+  8H2SO4= 

K2SO4+  2MnSO4+  5Fe2(SO4)3  +  8H2O 
2KMnO4  +  5H2C2O4+  3H2SO4  = 

K2SO4  -f  2MnSO4  +  ioCO2  +  8H2O. 

We  see  that  the  same  quantity  of  KMnO4  is  required 
for  5Fe2(SO4)2  as  for  5H2C2O4;  5Fe2(SO4)2  contains  10 
Fe  so  that  as  regards  KMnO4,  10  X  atomic  wt.  of  Fe  = 
5  X  Mol.  wt.  H2C2O4or  we  have,  Fe  standard  (o.oi  12)  : 
oxalic  acid  standard  :  :  10  X  56.02  :  5  X  90.03  or 
o.oi 1 2  :  x:  :  560.2  :  450.2.  x  =  0.0090. 

All  these  steps  are  not  necessary  however,  all  we 
need  do  is  to  consider  the  strength  of  N/6  oxalic  acid, 
then  this  will  be  equivalent  c.c.  perc.c.  with  N/5  KMnO4. 
From  the  reaction  we  see  that  one  H2C2O4  requires  one 
O  :  therefore  the  amount  necessary  to  bring  into  re- 
action eight  grams  of  oxygen  will  be  one  half  the  mole- 
cular weight  in  grams  per  liter  and  for  N/5  one  tenth  the 
molecular  weight  or  9.003  grams,  so  that  i  c.c.  will  be 
equal  to  0.0090  gram. 

To  calculate  the  standard  to  CaO  or  Cd,  we  shall 
also  have  one  tenth  the  molecular  or  atomic  weights,  be- 
cause these  are  precipitated  as  oxalates  whose  composi- 
tion is  CaC2O4  and  CdC2O4  and  the  oxalic  acid  deter- 
mined ;  so  that  one  molecule  of  oxalic  acid  corresponds 
to  a  molecule  of  CaO  or  an  atom  of  Cd. 

To  calculate  the  strength  of  N/5  KMnO4  in  terms  of 
molybdic  oxide  we  have  the  oxidation  of  Mo2O3  to 
MoO.3  or  two  MoO3  require  3  oxygen,  and  we  have 

288  :  x:  :  48  :  8.     x  =48. 

Therefore  N  MoO3  would  have  one  third  the  molecular 
weight  in  grams  per  liter  (48  grams)  and  N/5  one  fif- 
teenth (9.6  grams)  per  liter  or  i.c.c.  =  0.0096  gram 
MoO3. 


62  PERMANGANATE  CALCULATIONS. 

The  determination  of  phosphorus  by  KMnO4  is  in- 
direct for  as  the  ratio  of  MoO3  to  P  in  the  ammonium 
phospho-molybdate  is  12  to  i,  for  every  I44X  12  grams 
of  MoO3  found  by  titration  there  must  be  3 1  grams  of 
phosphorus  present:  so  we  have  MoO3  standard  :  P 
standard  as  1728  :  31  or  i  c.c.  N/5  KMnO4  =0.0001722 
gram  of  phosphorus. 

To  calculate  this  value  for  a  standard  KMnO4  solution 
not  normal  when  the  iron  standard  is  known :   we  have 
6KMnO4  =  ioMoO3  and 

6KMnO4  =  30  Fe  hence 

10  MoO3  correspond  to  30  Fe  and  the  Fe  standard: 
MoO3  standard  :  :  1680  :  1440  *  or  as  7:6  and  as  the 
phosphorus  standard  is  1.794%  of  the  MoO3  standard, 
the  phosphorus  standard  will  be  6/7  of  1.794%  of  the  Fe 
standard  or  the  Fe  standard  times  0.01538. 

A  rather  confusing  calculation  is  met  in  the  Ford- 
Williams  method  for  manganese  :  here  the  manganese 
is  precipitated  as  hydrated  MnO2  and  its  oxidizing 
power  measured.  Suppose  we  take  for  analysis  one 
gram  of  pig  iron,  and  to  the  precipitate  of  MnO2  add 
25  c.c.  of  N/10  Fe2(SO4)2  and  then  titrate  the  Fe2(SO4)2 
remaining  unoxidized  by  N/10  KMnO4,  using  12  c.c. 
What  is  the  percentage  of  manganese  ? 

The  reactions  are 
MnO2  +  Fe2(SO4)2  +  2H2SO4  = 

MnSO4  +  Fe2(SO4)3  +  2H2O. 

and  the  usual  reaction  between  the  excess  of  ferrous 
sulphate  and  permanganate.  The  solutions  used  were 
both  N/10and  therefore  correspond  c.c.  for  c.c.,  so  that  the 
oxidation  done  by  the  MnO2  was  equal  to  that  which 

*  According  to  the  latest  atomic  weights  this  ratio  should  be  1680.6  :  1439.9. 


PERMANGANATE   CALCULATIONS.  63 

would  have  been  done  by  13  c.c.  of  N/10  KMnO4.  Now 
we  have  2KMnO4  =  icFe 

iMnO2     =    2Fe 

Hence  2KMnO4  are  equivalent  to  5MnO2  or  5Mn,  or  Fe 
standard  :  Mn  standard  :  :  560.2  :  274.95  and  the  per- 
centage of  manganese  will  be  13  (number  of  c.c  of 
KMnO4  not  required) 

X  0.00275  -^  i  X  ioo  or  3.57%  Mn. 

Oxalic  acid  may  be  used  in  the  same  method  instead 
of  ferrous  sulphate  to  measure  the  oxidizing  power  of  the 
precipitate  and  the  excess  titrated  by  permanganate. 
Suppose  one  gram  of  spiegel  is  taken  for  analysis  and 
that  we  have  a  KMnO4  solution  (not  systematic)  whose 
strength  is  i.c.c.  =  0.006  grams  of  iron.  We  make  up 
a  solution  of  oxalic  acid  and  find  that  it  requires  25  c.c. 
of  the  KMnO4  solution  to  oxidize  20  c.c.  of  the  oxalic 
acid  solution.  To  the  precipitate  (hydrated  Mn(X) 
from  one  gram  of  spiegel,  80  c.c.  of  the  oxalic  acid  solu- 
tion are  added  and,  after  the  complete  solution  of  the  pre- 
cipitate, the  excess  is  titrated  by  KMnO4,  this  requires 
1 2. c.c.  To  calculate  the  percentage  of  manganese  :  If 
there  had  been  no  MnO3  present  the  80  c.c.  of  oxalic 
acid  would  have  required  100  c.c.  of  KMnO4,  but  only 
1 2  were  needed,  therefore  the  oxidation  done  by  the  MnO2 
present  was  equal  to  that  of  88  c.c.  of  KMnO4.  Now 
the  manganese  standard  of  our  KMnO4  is 

0.006  :  x  :  :  560.2  :  274.95     or  0.002945 
and     88  X  0.002945  -f-    i  X  100  =  25.92  the 
percentage  of  manganese. 

The  peculiar  part  of  this  method  is  that  it  is  the 
amount  of  KMnO4  not  used  which  is  the  basis  of  calcu- 
lation. 


64  IODINE   CALCULATIONS. 

IODINE    CALCULATIONS. 

Tenth  normal  iodine  solution,  12.685  grams  per  liter, 
can  be  used  for  a  great  many  titrations.  For  example, 
in  the  determination  of  H2S  and  so  of  sulphur  in  pig 
iron,  &c.  The  reaction  is 

H2S  +  2!  =  S  +  2HI. 

Five  grams  of  pig  iron  evolved,  on  treatment  with 
hydrochloric  acid,  H2S,   which  when  titrated  in  a  dilute 
solution  with  starch  as  indicator  required  6.2  c.c.  of  N/10 
iodine  solution.     What  is  the  percentage  of  sulphur  ? 
N/10  I  =  0.012685  gram  I  per  c.c.  =  0.001603 
gram  S  per  c.c.         6.2  x  0.0016  x  IQQ  =  o  I9g%  S> 

Iodine  reacts  with  H2SO3  as  follows : 

I2  +  H2SO3  +  H2O  =  2HI  +  H2SO4 
and  with  Na2S2O3  according  to  the  reaction, 
I2  +  2Na2S2O3  =  2NaI  +  Na2S4O6 

we  see  therefore  that  one  iodine  is  equivalent  to  yz  SO2, 
%  H2SO3  or  Y*  Na2SO3  and  to  one  Na2S2O3. 

The  latter  reaction  can  be  employed  for  the  deter- 
mination of  copper  as  follows : 

2Cu  (C2H3O2)2  +  4KI  =  Cu2I2  -f  I2  +  4KC2H3O2 
and  taking  the  last  two  reactions  we  see  that  one  mole- 
cule of  Na2S2O3  equivalent  to  one  iodine  is  equivalent 
to  one  copper,  so  if  we  have  a  N/]0  iodine  solution  we  can 
use  it  to  standardize  a  sodium  thiosulphate  solution  and 
so  indirectly  determine  copper  by  measuring  the  iodine 
liberated. 

Suppose  we  have  an  iodine  solution  of  unknown 
strength  and  a  potassium  dichromate  solution  whose 
strength  is  i  c.c.  =  0.0056  gram  of  iron.  To  find  the 
strength  of  our  iodine  solution  in  terms  of  sulphur. 


IODINE   CALCULATIONS.  65 

Make  up  a  sodium  thiosulphate  solution  of  any  con- 
venient strength  say  24.8  grams  of  the  crystallized  salt 
(Na2S2O3.  5H2O)  per  liter,  approximately  N/10. 

Dissolve  one  gram  of  potassium  iodide  in  water, 
acidify  strongly  with  HC1  and  add  25  c.c.  of  the  dichro- 
mate  solution.  Iodine  will  be  liberated  according  to  the 
reaction 


or  the  amount  of  K2Cr2O7  which  oxidizes  6Fe  liberates 
61  ;  therefore  the  strength  of  the  K2Cr2O7  solution  is 
i.  c.c.  =  0.012685  gram  of  iodine  and  the  weight  of 
iodine  liberated  will  be  25  X  0.012685  or  0.3171  gram. 

Into  this  we  run  the  sodium  thiosulphate  solution  till 
it  gives  no  blue  color  with  starch  ;  suppose  this  requires 
25.5  c.c.  then  i.c.c.  =  0.01243  gram  of  iodine.  Now 
take  any  convenient  amount  of  the  unknown  iodine 
solution  as  25  c.c.  and  titrate  it  with  the  sodium 
thiosulphate  solution;  suppose  it  requires  12.2  c.c.  then 

its    strength    in  terms    of  iodine    is    I2-2  x  0*01243  or 

D 

0.006058  and  against  sulphur  when  titrated  as  H2S, 

0.006058  :  x  :  :  126.85  :  16.035 
x  or  i  c.c.  =  0.000765  gram  sulphur. 
Arsenic  is  determined  by  iodine  according  to  the  re- 
action : 

Na3AsO3  +  I2  +  2NaHCO3  = 
Na3As04  +  2NaI  +  H2O  +  2CO2 

from  which  it  is  evident  that  2!  =  lAs  or  i  c.c.  N/10 
I  =  0.003754  gram  of  arsenic  or  0.00495  gram  of 
As4O6.  The  available  chlorine  in  bleaching  powder  can 
be  determined  according  to  a  similar  reaction  by  using 
a  solution  of  Na3AsO3,  which  can  conveniently  be 
standardized  by  the  N/10  iodine. 


66  ADJUSTMENT   OF    SOLUTIONS. 

Antimony  may  be  determined  by  sodium  thiosulph- 
ate  in  a  hydrochloric  acid  solution  as  follows : 
SbCl5+5KI  =    SbI3+I2+5KCl 
I2  +  2Na2S2O3  =  2NaI  +  Na2S4O6 

Suppose  we  have  a  solution  of  Na2S2O3  whose 
strength  is  i  c.c.  =  0.005  gram  Cu.  What  is  its 
strength  in  terms  of  antimony  ? 

We  see  by  comparing  the  reactions  that  2Cu  =  I2  and 
iSb  =  I2  hence 

0.005  :  x  :  :  63.6  :  60.215.     x  =  0.00473  gram. 

If  the  sodium  thiosulphate  were  N/10  we  should  see  at 

once  its  strength  against  copper  and  antimony,  for  one 

atom  of  copper  brings  into  reaction  one  iodine  equivalent 

to  one  half  atom  of  oxygen  or  a  N/10  solution  of  Cu  = 

6.36  grams  per    liter;    while    i    Sb  =   2!  =  i   oxygen. 

So  N/10  Sb  =   120.43  -^-  (2  x  10)   =  6.0215   grams  and 

i  c.c.  N/10  Na2S2O3    =   0.00636   gram   Cu 

=   0.00602    gram  Sb. 

ADJUSTMENT    OF    SOLUTIONS. 

As  this  is  in  no  sense  a  book  on  volumetric  analysis 
a  discussion  of  the  merits  of  the  different  liters  in  practi- 
cal use  is  out  of  place  and  the  calculations  are  given  so 
as  to  apply  to  whatever  unit  is  taken  for  the  volumetric 
apparatus. 

Suppose  we  have  made  up  ten  liters  of  sulphuric 
acid  approximately  normal  and  find  on  standardizing  that 
the  strength  is  i  c.c.  =  0.050  gram  H2SO4  and  that  in 
making  the  test  we  have  used  50  c.c.  of  the  solution. 
We  have  left  9,950  cubic  centimeters  and  the  solution  is 
too  strong.  To  find  the  amount  of  water  to  add  to  make 
it  normal :  let  x  be  the  volume  required  so  that  the 
solution  shall  be  normal,  then  x  times  0.049043  is  the 


ADJUSTMENT   OF   SOLUTIONS.  67 

weight  of  H2SO4  contained  therein  ;  9,950  X  0.050  is  the 
weight  of  H2SO4  in  our  solution.  These  quantities  are 
equal  as  they  will  not  be  affected  by  dilution  therefore 
we  have 

9,950  X  0.050  =  x  X  0.049043. 

x  =  10,144.1  c.c.  So  the  amount  of  water  to  be  added  is 
10,144.1  -  9,950  or  194.1  c.c.  (These  cubic  centimeters 
must  accord  with  the  liter  adopted). 

Suppose  the  solution  gave  on  standardizing  i  c.c.  = 
0.048  gram  H2SO4.  The  solution  is  too  weak;  by  a 
similar  method  we  can  find  the  amount  of  water  in  excess 
and  then  add  the  proper  quantity  of  sulphuric  acid,  we 
have 

9950  X  0.048  =  x  X  0.049043.     x  =  9738.4. 
The  excess  of  water  is  therefore  9950  -  9738.4  or  216.6 
c.c.     For  the  amount  of  sulphuric  acid  to  be  added  see 
Chapter  VII. 

It  will  be  readily  seen  that  it  is  much  more  convenient 
to  make  up  the  solutions  too  strong  rather  than  too 
weak. 

If  we  have  a  solution  of  say  Na2CO3  which  is  too  weak ; 
find  out  the  amount  of  water  in  excess,  then  calculate  the 
amount  of  the  salt  for  this  volume  of  water  and  add  it. 
The  change  in  volume  due  to  the  solution  of  the  salt  can 
be  neglected. 

Example.  A  solution  of  Na2CO3  was  found  to  have 
a  strength  i  c.c.  =  0.047  gram  of  H2SO4.  How  much 
Na2CO3  must  be  added  to  two  liters  to  make  it  normal? 

2000  X  0.047  =  x  X  0.049043.     x  =  1,916.6. 
There  are  83.4  c.c.  of  water  in  excess.     Now,  a  normal 
solution  of  Na2CO3   contains   YZ    molecular  weight  in 
grams  per  liter  or  53.055  grams,  so  the   amount  to  be 
added  is 

53-°55  :  Y  '•  :  Jooo  :  83.4.     y  =  4.4244  grams. 


68         EXCESS   NECESSARY   TO   AFFECT   THE    INDICATOR. 

We  may  have  two  standard  solutions  of  a  reagent 
and  desire  to  make  a  third  of  a  certain  intermediate 
strength  —  for  example  suppose  we  have  two  solutions  of 
NH4SCN 

(a)  i  c.c.  =  0.01275  gram  of  silver 

(b)  i  c.c.  =  0.00447  gram  of  silver 

and  we  wish  to  make  one  liter  of  a  solution  equivalent 
to  10  mgs.  of  silver  per  c.c.  How  much  of  each  shall 
we  mix.  Let  x  be  the  number  of  c.c.  of  (a)  ;  then  1000  — 
x  =  number  of  c.c.  of  (b)  :  and  we  have 

x  (0.01275)  +  (IOQO-X)  0.00447  =  10 
(as  one  liter  will  contain    10  grams  if   i   c.c.  contains 
10  mgs.)     This  gives  667.9  c.c.  of  (a)  and  332.1  c.c.  of 

(b). 

If  we  have  a  solution  of  KMnO4  containing  2  grams 
per  liter  and  a  N/5  solution  of  the  same  to  make  two 
liters  of  a  N/10  KMnO4.  Let  x  be  the  volume  of  the  2 
grams  solution  in  liters  and  (2  —  x)  the  volume  of  the  N/5 
solution  ;  then  2  x  +  (2  —  x)  6.324  =  2  (3.162)  and  x  = 
i. 462 5  liters,  (2-x)  =  0.5375  liter- 

SUBTRACTION    OF    EXCESS    NECESSARY   TO    AFFECT  THE 
INDICATOR. 

In  some  titrations,  conspicuously  that  of  zinc  by  potas- 
sium ferrocyanide  and  that  of  lead  by  ammonium  molyb- 
date,  the  end  point  is  not  very  sharp  and  a  considerable 
excess  of  the  standard  solution  is  used  before  the  indica- 
tor is  affected.  This  amount  should  be  determined  by 
running  a  blank  test  under  the  same  conditions  as  to 
temperature,  bulk,  acidity  &c.,  and  should  be  subtracted 
from  the  burette  reading  before  calculating  the  results  ; 
for  this  amount  is  the  excess  necessary  to  affect  the  indi- 
cator and  is  required  when  there  is  no  metal  present. 
Other  conditions  remaining  the  same  this  excess  is  usu- 


EXCESS    NECESSARY   TO   AFFECT   THE   INDICATOR.          69 

ally  proportional  to  the  bulk,  but  has  to  be  determined 
for  each  set  of  conditions. 

This  subtraction  of  the  excess  necessary  to  turn  the 
indicator  can  be  omitted  when  the  solution  titrated  con- 
tains nearly  the  same  weight  of  metal,  as  well  as  the 
same  volume  &c.,  as  was  used  in  standardizing,  for  then 
this  appears  in  the  lower  standard  of  the  solution  and  no 
appreciable  error  is  introduced.  These  conditions  are 
however  almost  impossible  to  attain  with  a  variety  of 
ores,  so  that  the  best  plan  is  to  make  the  subtraction 
from  every  reading,  for  a  color  and  set  of  conditions 
found  satisfactory. 

The  following  will  illustrate  this  fact — 

A  solution  of  crystallized  potassium  ferrocyanide  was 
made  up  for  the  titration  of  zinc  ores,  approximately,  44 
grams  per  liter.  To  standardize ;  200  mgs.  of  pure  zinc 
were  titrated  under  the  standard  conditions,  using  uran- 
ium acetate  as  an  indicator.  Then  a  blank  was  run 
under  the  same  conditions. 

0.200  gram  of  zinc -f  H2O,  HC1  &c  required  20.45  c-c- 
Blank  (HgO,  HC1  &c)  required  0.45  c.c. 
0.200  gram  of  zinc  required  20  c.c. 
ic.c.  =  o.oio  gram  of  zinc. 

Disregarding  the  blank  the  strength  becomes 
i. c.c.  =  0.00978  gram  of  zinc. 

We  now  titrate  one  gram  of  an  ore  and  use  21  c.c.  : 
using  the  blank  this  gives  20.55%:  disregarding  the 
blank  and  using  the  lower  standard  gives  20.538%. 
Results  close  enough  for  ordinary  work.  But  suppose 
we  now  titrate  a  rich  ore,  say  one  gram  requires  un- 
der the  same  conditions  56  c.c.  :  using  the  blank  we 
get  55-55%  and,  disregarding  the  blank  and  using  the 
lower  standard,  we  get  54.768%.  This  involves  an 
error  not  to  be  disregarded  under  any  circumstances. 


EXAMPLES. 


1.  Calculate  the  number  of  grams  per  liter  to   give 
normal  solutions  of  oxalic  acid,  of  tartaric  acid,  of  acetic 
acid  and  of  citric  acid. 

Ans.  H2C2O4,  45.01  grams;  H2C4H4O6,  75.024 
grams;  HC2H3O2,  60.032  grams;  H3C6H5O7,  64.021 
grams. 

2.  Calculate  the  number  of  grams  per  liter  to  give 
half  normal    solutions  of  I,  of  Na2S2O3,  5H2O,   and  of 
SO2. 

Ans.  I,  63.425  grams;  Na.3S2O3,5H2O,  124.16 
grams;  SO2,  16.017  grams. 

3.  Calculate    the    amounts    necessary  to  make  two 
liters  of  N/10  NaCl  and  NaBr. 

Ans.     NaCl  11.7  grams;     NaBr  20.6. 

4.  Calculate  the  strength  of  N/go  KMnO4  to  be  used 
in  titrating  K4Fe  (CN)6. 

Ans.     1.581  grams  per  liter. 

5.  What  is  the  strength  of  N/10  KMnO4  in    terms  of 
Fe?  Fe203?  Fe3O4?  FeSO4,7H2O? 

Ans.     0.0056.    0.0080.    0.007735.    0.02782. 

6.  What  is  the  strength  of  N/10  KMnO4  in  terms  of 
H8C2O4?  ofCaO?  ofCaCO3?  ofCaSO4? 

Ans.     0.0045.      0.0028.     0.0050.     0.0068. 

7.  What  is  the  strength  of  N/10  KMnO4in  terms  of 
P  and  of  MoO3  according  to  Noyes  method? 

Ans.     0.0000861  and  0.0048. 

(70) 


EXAMPLES.  71 

8.  What  is  the  strength  of  a  N/10  sodium  thiosulphate 
solution  in  terms  of  iodine  ?  of  copper  ? 

Ans.     0.012685.  0.00636. 

9.  What  is   the    strength  of  a  N/5  iodine  solution  in 
terms  of  SO2?  H2SO3?  Na2SO3? 

Ans,     0.0064.     0.0082.     0.012616. 

10.  How    many  c.c.  of  N/5  Na2S2O3  solution  will  be 
required  to  react  with  the  iodine  liberated  by  20  c.c.  of  a 
N/10  K2Cr2O7  solution  ?  Ans.      10  c.c. 

1 1.  How  many  grams  of  copper  will  give  when  pre- 
cipitated by  potassium  iodide  sufficient  iodine  to  require 
20  c.c  of  N/10  SO2   solution?      How  many  for  10  c.c.  of 
•N/io  Na2S2O3  solution?        Ans.     0.1272  and  0.0636. 

1 2.  Five  grams  of  pig  iron  took  1 2  c.c.  of  N/100  iodine 
solution.     What  is  the  percentage  of  sulphur  ? 

Ans.     0.0384%. 

:3-    1'5  grams  of  pig  iron  took  78  c.c.  of  N/10  KMnO4. 
What  is  the  percentage  of  phosphorus  ? 

Ans.  0.447%. 

14.  In   determining  iron  in  an  ore,  by  mistake  too 
much  N/10  K2Cr2O7  was  run  in :  the  burette  read  45  c.c.  : 
20   c.c.  of  N/10  Feo(SO4)2  were  added  and  the  titration 
finished:   the  burette  then  read  57  c.c.    How  much  dichro- 
mate  was  required  for  the  iron  ore?         Ans.     37  c.c. 

15.  Two  liters  of  KMnO4  solution  were  made  up 
approximately  N/5.      It  was  found  on  standardizing  that 
the  strength  was  i.c.c.  =0.01175  gram  Fe  and  48.2  c.c. 
were   used.     How  much  water  must  be  added  to  what 
remains  to  make  it  N/5?  Ans.     95.8  c.c. 

1 6.  How  much  NaOH  must  be  added  to  1.890  liters 
of  NaOH   solution,   whose  strength  is   i  c.c.   =  0.045 
gram  H2SO4,  to  make  it  normal?    How  much  water  to 
makeitN/2? 

Ans.    ,6.172  grams  NaOH.      1581.3  c.c.  water. 


72  EXAMPLES. 

1 7.  How  many  grams  per  liter  of  KMnO4  will  give 
a  solution  of  such  strength  that  i  c.c.  =  i%  of  iron  when 
o.  50  gram  is  taken  for  analysis  ? 

Ans.     2.8222  grams. 

1 8.  How  many  grams  per  liter  of  K2Cr2O7  will  give 
a  solution  of  such  strength  that  i  c.c.  =  i%  of  iron  when 
0.60  gram  is  taken  for  analysis  ? 

Ans.     5.2570  grams. 

19.  How  many  grams  of  Na2S2O3,  5H2O  per  liter 
will  give  a  solution  such  that  i  c.c.  =  0.5%  Cu,  when 
one  gram  is  taken  for  analysis  ? 

Ans.      19.522  grams. 

20.  How  much  water  must  be  added  to  two  liters  of 
KMnO4,  i  c.c.  =  0.0065  gram  iron,  to  make  it  N/10? 

Ans.     321.4  c.c. 

21.  How  much  water  must  be  added  to  three  liters 
of  K4Fe(CN)6,    i  c.c.  =0.0115   gram  zinc,  to  make  it 
read  percentage  directly  when  one  gram  is  taken  for 
analysis?  Ans.     450  c.c. 

22.  Given  a  solution  of  K2Cr2O7,    i  c.c.   =  0.0042 
gram  Fe,  how  many  grams  per  liter  of  KMnO4  will  give 
a  solution  of  equal  strength  ? 

Ans.     2.3715  grams. 

23.  One   gram  of  an    ore  containing   arsenic   was 
treated   according  to  Pearce's   method  and   the  arsenic 
precipitated  as  Ag3AsO4 :   the  silver  so  combined  was 
titrated  by  NH4SCN.      The  amount  used  was  27.9  c.c. 
and  the  strength  of  the  solution  i.c.c.  =  0.007  gram  of 
silver.     What  was  the  percentage  of  arsenic  ? 

Ans.     4.52%. 

24.  A  solution    of  Na2S2O3  was   made    containing 
about  40  grams  of  the  crystallized  salt  per  liter ;   this  was 
standardized  by  portions  of  copper  0.200  gram  each; 


EXAMPLES.  73 

they  required    19.6  c.c.  :  0.50  gram  of  a  copper  matte 
took  37.6  c.c.     What  is  the  percentage  of  copper? 

Ans.      76.73%. 

25.  Given    a  solution    of  KMnO4   i  c.c.    =  0.0056 
gram  of  iron.      What  is  its  strength  in  terms  of  manga- 
nese by  the  Volhard  method  and  by  the  Ford- Williams 
method? 

.  (  Volhard  i  c.c.  =  0.00105  gram. 

ns'      \  Ford-Williams  i  c.c.  =  0.00275  gram. 

26.  The  MnO2  from  one  gram  of  spiegel  was  dis- 
solved  in    1.5    grams   of  ferrous   ammonium    sulphate, 
(NH4)2SO4,   FeSO4,  6H2O,   and    then    titrated  by  N/10 
KMnO4  :  2.05  c.c.  were  used.   What  is  the  percentage  of 
manganese?  Ans.    9.94%. 

27.  In  one  gram  of  an  antimony  alloy,  the  antimony 
was  determined  by  N/10  Na2S2O3.  24.2  c.c.  were  required. 
What  is  the  percentage  of  antimony  ? 

Ans.      i4.57%- 

28.  What  amount  of  type  metal  must  be  taken  for 
analysis    so  that  the  burette  shall  read  percentage   of 
antimony  when  N/10  Na2S2O3  is  used  ? 

Ans.     0.602  gram. 

29.  Five  grams  of  bleaching  powder  were  mixed  in 
a  mortar  with  water  and  diluted  to  one  liter  :    50  c.c.  of 
this  required  30  c.c.  of  N/10  Na3AsO3  solution.     What  is 
the  percentage  of  available  chlorine  ? 

Ans.     42.54%. 

30.  Under  the  conditions  of  the  preceding  example 
what  fraction  of  normal  will  read  percentage  direct  ? 

Ans.     _*L 

I4.I8 

31.  To  50  c.c.  of  a  solution  of  chlorine  an  excess  of 
potassium  iodide  was  added  ;  the  liberated  iodine  was 
then  estimated  by  a  N/10  solution  of  Na2S2O3  using  starch 


74  EXAMPLES. 

as   an   indicator;    22.5    c.c.    of   Na2S2O3  solution  were 
used.     What  is  the  strength  of  the  chlorine  solution  ? 
Am.     0.001595  gram  per  c.c. 

32.  Twenty  grams  of  an  adulterated  sample  of  vin- 
egar,   freed  from   soluble    sulphates,    required    16   c.c. 
N    NaOH  :    10   grams  gave  0.3162    gram  of  BaSO4. 
What  is  the  percentage  of  acetic  acid?    Ans.     3.176%. 

33.  0.4  gram  of  impure  caustic  soda  required  when 
titrated  directly  95  c.c.  of  N/10  HC1:    another  0.4  gram 
was   dissolved,    an    excess  of  barium    chloride  added, 
and   then   one-half  titrated,   requiring   46    c.c.      What 
are  the  percentages  of  NaOH  and  Na2CO3? 

A  \  NaOH  92.13%. 

\  Na2C03  3-98%. 

34.  Two  grams  of  impure  potassium  bicarbonate  re- 
quired when  titrated  cold,  in  a  dilute  solution,  6  c.c.  of 
N/10  H2SO4:  two  grams  when  titrated  boiling  hot  re- 
quired 1 8  c.c.  of  N  H2SO4.     What  are  the  percentages 
ofK2CO3andof  KHCO3? 

Ans.     4. 14%  and  84.09%. 

35.  What  volumes  of  N/2  and  of  N/10  H2SO4  must  be 
mixed  to  give  two  liters  of  N/5  H2SO4? 

Ans.     y2  liter  N/2  and  i^  liters  N/10. 
36.  A  N/10  iodine  solution   is   2%  too  strong.     How 
many  cubic  centimeters  of  a  N/100  iodine  solution  must  be 
added  to  one  liter  to  make  it  right  ? 

Ans.     22.13  c-c- 

37.  Under  certain  conditions,  in  the  titration  of  lead 
by  ammonium  molybdate,  0.75  c.c.  was  found  necessary 
to  turn  the  indicator:  0.300  gram  of  lead  took  32  c.c. 
of  the  solution  (without  allowance).  If  the  subtraction 
of  the  blank  is  omitted,  what  is  the  error  in  the  titration 
of  one  gram  of  a  60%  ore  ?  A  ns.  o.  7 1  %. 


EXAMPLES.  75 

38.  How  much  more  N  NaOH  will  it  take  to   neu- 
tralize i  gram  of  HC1  than  i  gram  HBr? 

Ans.     15.08  c.c. 

39.  How  much  more  N  NaOH  will  it  take  to  neu- 
tralize i  gram  of  NaHSO4  than  N  KOH  to  neutralize  i 
gram  of  KHSO4?  Ans.     0.99  c.c. 

40.  0.2    gram    of  a  nitrogenous  organic  compound 
was  heated  with  soda-lime  and  the  NH3  evolved  caught 
in  50  c  c.  N/10  HC1:  the  excess  of  HC1  was  neutralized 
by  14  c.c.  of  N/s  NaOH.     What  was  the  percentage  of 
nitrogen?  Ans.      15.44%. 


CHAPTER    VII. 

CALCULATIONS   OF   DENSITY   OF    SOLIDS   AND   LIQUIDS. 
DENSITY    OF    SOLIDS. 

THE  relative  density  or  as  it  is  commonly  called 
specific  gravity  of  solids  and  liquids  is  compared  with 
that  of  water  at  the  temperature  of  maximum  density  as 
unity.  So  if  we  say  a  substance  has  a  specific  gravity 
of  2.5,  we  mean  that  a  given  volume  has  2.5  times  the 
weight  of  the  same  volume  of  water  at  4°C  ;  or  that  its 
density  relative  to  that  of  water  is  as  2.5  is  to  i. 

The  density  of  solids  can  be  determined  by  the  fol- 
lowing methods.  The  usual  classification  is  given  here 
but  no  formulae,  as  these  are  more  confusing  than  help- 
ful and  do  not  lead  to  a  thorough  understanding  of  the 
subject. 

A.  Solids  heavier  than  water  and  insoluble, 
(i)  Weigh  in  air  and  then  in  water. 

According  to  the  law  of  Archimedes,  a  body  im- 
mersed in  a  fluid  loses  a  part  of  its  weight  equal  to  that 
of  the  fluid  it  displaces.  So  the  weight  of  the  solid  in 
air  minus  its  weight  in  water  gives  the  weight  of  an 
equal  volume  of  water;  as  the  volume  of  water  dis- 
placed is  equal  to  that  of  the  object,  unless  it  is  porous, 
when  it  must  be  allowed  to  soak  in  water  for  some  time  to 
expel  the  air.  So  if  we  divide  the  weight  in  air  by  the 
loss  in  weight  in  water,  we  have  the  specific  gravity  or 
density  compared  to  that  of  water  at  the  temperature  at 
which  the  experiment  was  carried  out.  This  can  then 

(76) 


DENSITY   OF   SOLIDS.  77 

if  necessary  be  corrected  for  the  expansion  of  water  and 
the  true  specific  gravity  compared  to  water  at  4°C  ob- 
tained. For  great  accuracy  the  weight  in  air  should  be 
reduced  to  weight  in  vacuo. 

For  example : —  Suppose  we  have  a  piece  of  brass 
whose  weight  in  air  is  24.076  grams  and  in  water 
21.2436  grams.  The  weight  in  air  divided  by  the  loss 
in  weight,  2.8324  grams,  gives  the  specific  gravity,  8.5, 
referred  to  that  of  water  at  the  temperature  of  the  ex- 
periment. 

(2)  If  the  substance  is  in  small  pieces  or  is  a  powder 
the  following  method  is  used: 

(a)  Weigh  the  substance  in  air. 

(b)  Weigh  a  small  flask  filled  either  entirely  with 
water  or  to  a  certain  mark. 

(c)  Place  the  substance  in  the  flask,  remove  air,  if 
necessary,  by  boiling  and  weigh  with  the  water  at  the 
same  point  as  before  (either  filled  or  to  the  mark). 

It  is  evident  that  a  volume  of  water  equal  to  that  of 
the  substance  has  been  displaced :  to  get  the  weight  of 
this  volume  of  water;  weight  of  substance  (a)  +  weight 
of  flask  and  water  (b)  —  weight  of  substance  in  flask 
and  water  (c)  =  weight  of  water  displaced.  Then  weight 
of  substance  in  air  divided  by  the  weight  of  an  equal 
volume  of  water  gives  specific  gravity.  For  example  : 

Weight  of  powder,  7  grams. 

Weight  of  flask  and  water  (to  mark)  55  grams. 

Weight  of  flask,  powder  and  water  (to  mark)  59 
grams. 

We  have   55  +   7  —  59  =  3  weight   of  volume   of 

water  displaced  by  powder:  and  —=2.333,  tne  specific 
gravity  of  the  powder. 


78  DENSITY  OF   SOLIDS. 

By  having  graduations  on  the  neck  of  the  flask,  the 
volume  of  the  substance  may  be  read  off  by  the  rise  of 
the  water  in  the  flask,  then  this  volume  in  c.c.  multiplied 
by  the  weight  of  i  c.c.  of  water  at  the  temperature,  gives 
the  weight  of  an  equal  volume  of  water. 

B.  Solids  lighter  than  water  and  insoluble. 

As  these  solids  are  lighter,  they  cannot  be  weighed 
in  water  without  a  sinker. 

The  data  necessary  is — 

Weight  of  substance  in  air. 

Weight  of  sinker  in  water. 

Weight  of  substance  and  sinker  in  water. 

Or  if  we  have  the  weight  of  sinker  in  air  and  its 
specific  gravity  we  can  calculate  the  weight  of  the  sinker 
in  water. 

As  the  substance  is  lighter  than  water,  the  weight 
of  sinker  and  substance  in  water  is  less  than  the  sinker 
alone. 

If  we  subtract  from  weight  of  sinker  and  substance 
in  water  the  weight  of  sinker  in  water,  we  have  the 
weight  of  substance  in  water,  a  negative  quantity  and 
the  measure  of  its  buoyant  power. 

Now  the  weight  in  air  divided  by  the  weight  in  air 
minus  the  weight  in  water  is  the  specific  gravity,  but 
the  weight  in  air  minus  the  negative  weight  in  water 
equals  the  weight  in  air  plus  the  buoyant  power,  so  we 
have  in  this  case  the  specific  gravity  is  equal  to  the 
weight  in  air  divided  by  the  weight  in  air  plus  the  differ- 
ence in  weights  of  the  sinker  in  water  and  the  substance 
and  sinker  in  water. 

To  illustrate  by  an  example. — 

Weight  of  substance  in  air  4  grams. 

Weight  of  sinker  in  water  9  grams. 


DENSITY  OF  SOLIDS.  79 

Weight  of  substance  and  sinker  in  water  8  grams. 

Difference  in  weight  of  sinker  in  water  and  substance 
and  sinker  in  water  is  one  gram :  weight  in  air  plus  this 
difference  (or  minus  the  negative  loss  in  weight  of  one 
gram)  is  5. 

~  =  0.8  the  specific  gravity  of  the  substance. 

If  we  have  the  weight  of  the  sinker  in  air  and  its 
specific  gravity. 

The  weight  t  in  air  divided  by  the  specific  gravity 
gives  the  volume,  or  expressed  in  grams,  the  weight  of 
an  equal  volume  of  water.  This  is  equal  to  the  loss  of 
weight  in  water. 

The  weight  in  air  minus  the  loss  of  weight  in  water 
gives  the  weight  in  water. 

For  example,  suppose  the  sinker  weighed  1 2  grams 
and  its  specific  gravity  was  8. 

— =  1.5  (loss  in  weight)  and  12  —  1.5  or  10.5  =  weight  of 

8 

sinker  in  water. 

C    Solids  heavier  than  water  but  soluble. 

Weigh  in  air  then  in  a  liquid  of  known  specific 
gravity  in  which  the  substance  is  insoluble,  such  as  tur- 
pentine, petroleum,  alcohol,  &c. 

The  weight  in  air  divided  by  the  loss  in  weight  gives 
the  density  relative  to  the  liquid  used  ;  suppose  this  is 
7  and  the  specific  gravity  of  the  liquid  0.8,  to  find  the 
specific  gravity  we  have 

7  :  x  : :  i  :  0.8  or  7  x  0.8  =  5.  6  Sp.  Gr. 
Or,  the  loss  in  weight  of  the  solid  equals  the  weight  of 
liquid  displaced.     Let  this  be  x  and  y  the  weight  of  the 
same    volume   of  water;     then  x  :  y  :  :  0.8  :  i   and  the 
weight  in  air  divided  by  y  gives  the  specific  gravity. 


80  DENSITY  OF   SOLIDS. 

To  determine  the  specific  gravity  of  rock  salt.    Given  : 
Weight  of  sample  in  air  10.436  grams 
Weight  of  sample  in  turpentine  6.3547  grams 
Specific  gravity  of  turpentine  0.86 
10.436  -  6.3547  =  4.0813  loss  in  weight  in  turpentine 
equal  to  the  weight  of  an  equal  volume  of  turpentine. 
Let  y  =  weight  of  an  equal  volume  of  water 
4.0813  :  y  :  :  0.86  :  I.     y  =  4.7457  and 

—  lz~—  =  2.  1  09  Sp.  Gr.  of  rock  salt. 
4-7457 


or      *        =2.5572  density  relative  to  turpentine  and 
4.0813 

2.5572  :  x:  :  i  :  0.86.     x  =  2.199  Sp.  Gr.  of  rock  salt. 

D.   Solid  lighter  than  water  and  soluble. 

Weigh  with  a  sinker  in  some  liquid  which  does  not 
act  on  the  substance  and  calculate  the  density  relative 
to  the  liquid;  then  from  the  specific  gravity  of  the 
liquid  calculate  the  specific  gravity  as  under  C. 

To  determine  the  specific  gravity  of  lithium  — 

Weight  of  lithium  in  air   3.  grams 

Weight  of  sinker  in  air    10.  grams 

Weight  of  lithium  and  sinker  in  kerosene  7.9222 
grams. 

Specific  gravity  of  sinker  (silver)  10.5 

Specific  gravity  of  kerosene  used  0.829 

First  find  the  specific  gravity  of  the  sinker  referred 
to  kerosene 

0.829  :  i  :  :  10.50  :  x.     x  =  12.6658 

Next  find  the  weight  of  the  sinker  in  kerosene    we 

have  -  =  12.6658.      x  =  9.2104 

10-  x 


DENSITY   OF    LIQUIDS.  81 

Now  following  the  method  for  solids  lighter  than 
water  and  insoluble 

3 =  0.6996  density  of  lithium 

3  +  9.2104-7.9222 

relative  to  kerosene :  and  to  refer  to  water  we  have, 
0.829  :  i  :  :  x:  0.6996.     x  =  0.57996 

The  specific  gravity  of  lithium  is  therefore  0.58. 

In  these  examples  no  particular  attention  has  been 
given  to  the  temperature  of  the  water,  but  the  specific 
gravity  has  simply  been  referred  to  water  at  the  same 
temperature  (that  of  the  experiment).  To  refer  these 
specific  gravities  to  water  at  4°C;  look  up  in  the  table, 
at  the  back  of  the  book,  the  weight  of  a  cubic  centimeter 
of  water  at  the  particular  temperature  used,  then  substi- 
tute in  the  proportion, 
Sp.  Gr.x°C  :  Sp.  Gr.  4°C :  :  i  :  wt.  i  c.c.  water  at  x°C.  * 

This  is  the  same  proportion  used  for  the  conversion 
of  the  relative  density  of  a  solid  in  turpentine  to  density 
referred  to  water  (specific  gravity)  and  it  is  a  similar 
proposition  ;  for  water  at  say  20° C  is  a  liquid  lighter 
than  water  at  4°C  so  the  specific  gravity  referred  to 
water  at  4°C  will  necessarily  be  less. 

The  correction  for  the  weight  of  the  solid  in  air  is 
too  small  to  have  significance  here  and  will  be  discussed 
in  the  next  chapter. 

DENSITY    OF    LIQUIDS. 

The  specific  gravity  of  liquids  is  determined  as 
follows : 

ist.   By  the  pyknometer  or  specific  gravity  bottle. 

*  The  table  also  gives  the  volume  of  one  gram  of  water  at  different  tempera- 
tures. To  use  these  values  substitute  in  the  proportion  Sp.  Gr.  X°C  :  Sp.  Gr. 
4°C  :  :  vol.  at  X°C  :  1. 


82  DENSITY  OF   LIQUIDS. 

2d.  By  weighing  a  solid  of  known  specific  gravity  in 
the  liquid. 

3d.  By  hydrometers. 

4th.  By  Westphal's  balance.  This  is  really  only  a 
modification  of  the  second  method. 

ist.  In  the  pyknometer  we  can  compare  directly  the 
weights  of  the  same  volume  of  the  liquid  and  of  water  ; 
so  that  the  calculation  only  involves  the  subtraction  of 
the  weight  of  the  pyknometer  and  the  division  of  the 
weight  of  the  liquid  by  that  of  an  equal  volume  of  water. 

2d  If  we  have,  for  example,  a  piece  of  brass,  which 
weighs  in  air  10  grams,  in  water  8.824  grams  and  in  the 
liquid  of  unknown  specific  gravity  9  grams :  we  have  in 
each  case  the  loss  of  weight  of  the  brass  equals  the 
weight  of  the  volume  of  liquid  displaced  ;  and,  as  the 
volume  is  the  same  in  both  cases,  the  loss  in  weight  is 
directly  proportional  to  the  specific  gravity  and  we  have 

1.176        :  i  :  :  I         :          x.       x  =  0.85 

Loss  of  wt.     Loss  of  wt.     Sp.  Gr.  of  Sp.  Gr.  of 
in  water.        in  Liquid.          water.  liquid. 

Or  if  we  know  the  weight  of  the  brass  in  air  and  its 
specific  gravity  we  can  calculate  the  loss  of  weight  in 
water  as  already  explained  under  the  density  of  solids. 

3d  Hydrometers  are  made  either  to  read  specific 
gravity  direct,  like  the  lactometer,  or  are  graduated  ac- 
cording to  some  arbitrary  scale  of  degrees. 

Degrees  Beaume  and  specific  gravity  may  be  com- 
pared by  reference  to  the  tables  at  the  end  of  the  book. 

Degrees  Twaddell  have  a  direct  relation  to  specific 
gravity.  200°  represent  the  densities  between  i  and  2 
so  i  °  represents  a  difference  of  specific  gravity  of  0.005, 

.r  •      •         ,    N°  (Twaddell). 

hence  the  specific  gravity  is  i  -f i —       '- 

200 


DENSITY  OF   LIQUIDS.  83 

36°  Twaddell  is  1.18  Sp.  Gr. 

94°  Twaddell  is  i  .47  Sp.  Gr.  and  so  on. 

4th.  In  using  the  Westphal's  balance  each  weight 
gives  the  figure  for  a  decimal  place  beginning  with  the 
largest.  If  the  liquid  is  heavier  than  water  a  weight 
(largest  size)  must  be  placed  on  the  hook  to  help  sink 
the  float.  This  weight  gives  i  (before  the  decimal  point) 
and  the  remaining  weights  the  decimals. 

Under  the  specific  gravity  of  liquids  come  the  prob- 
lems of  dilution  to  a  certain  specific  gravity.  This  can 
of  course  be  done  without  any  calculation  by  the  use  of 
a  hydrometer.  The  following  method  will  however  be 
found  useful,  although  not  strictly  accurate  on  account 
of  a  small  contraction  of  volume  on  mixing  which  is 
found  by  very  accurate  measurements.  * 

Suppose  we  wish  to  make  200  c  c.  of  ammonia  water 
0.96  Sp.  Gr.  and  we  have,  ammonia  water  0.88  Sp.  Gr. 
and  water.  The  ratio  is  inversely  as  the  difference  in 
specific  gravities  .08  (.96  —  .88)  :  .04  (i.oo  —  .96)  :  : 
volume  of  water  :  volume  of  ammonia :  two  parts  of 
water  to  one  of  strong  ammonia,  or  66.66  c.c.  diluted 
to  200. 

Required  to  make  six  liters  of  nitric  acid  1.26  Sp.  Gr. 
having  nitric  acid  i  .42  Sp.  Gr.  and  water. 

26  :  1 6  :  :  volume  of  acid  :  volume  of  water  :  or  16 
parts  of  water  for  42  of  diluted  acid  :  for  6  liters,  16  :  42 
:  :  x  :  6.  x  =  2.285.  Or,  let  y  =  volume  of  nitric  acid 
in  liters  1.42  y  +  6  — y  =  6  (1.26).  y  =  3.714. 

These  methods  can  be  applied  to  any  case  of  mixing, 
as  the  contraction  previously  mentioned  will  not  intro- 
duce any  error  appreciable  in  ordinary  work. 

*  See  Wade,  Journal  Chemical  Society  p.  254,  1899. 


84  DENSITY  OF   LIQUIDS. 

Suppose  we  wish  to  make  up  two  liters  of  a  stan- 
dard sulphuric  acid  solution  containing  ten  grams  of 
H2SO4  per  liter.  Dilute  pure  concentrated  sulphuric 
acid  with  water  to  about  1.7  Sp.  Gr.  allow  to  cool  to 
i5°C  and  then  take  the  specific  gravity  most  accurately, 
suppose  it  to  be  1.689.  The  percentage  of  H2SO4  can 
be  found  by  referring  to  a  table*  to  be  76.29.  The  weight 

r          11-  -J  1.         j-i          J          MI     i          IO  X   2    X    IOO 

of  sulphuric    acid  to    be   diluted    will   be  — 

76.29 

=  26.2157  grams.  This  when  diluted  with  water  to  two 
liters  at  15°  C  will  give  a  solution  containing  ten  grams 
of  H2SO4  per  liter.  In  this  calculation  the  specific 
gravity  is  referred  to  water  at  15°  C  as  unity,  this  is  to 
agree  with  the  liter  used  in  dilution  (volume  of  1000 
grams  of  water  at  15°  C).  Should  we  wish  to  compare 
this  with  other  specific  gravities  it  can  be  reduced  to 
specific  gravity  at  4°  C  as  already  explained. 

This  brings  us  to  the  consideration  of  changes  in 
specific  gravity  due  to  temperature. 

With  solids  the  change  in  volume  is  so  small 
that  no  appreciable  error  is  introduced  by  neglecting 
this  variation  within  the  limits  of  laboratory  temper- 
ature. 

With  liquids  the  variations  of  specific  gravity  due  to 
changes  in  temperature  are  greater  and  cannot  be  dis- 
regarded. Unless  specially  stated  the  specific  gravity 
is  referred  to  water  at  4°  C  as  unity  and  for  accurate 
work  the  determination  must  be  carried  out  at  that  tem- 
perature. For  many  important  solutions  and  liquids  the 
corrections  for  temperature  have  been  worked  out  and 
are  to  be  found  in  the  books  on  technical  chemistry. 

*  Marshall,  J.,  Society  Chemical  Industry,  vol.  18,  p.  5,  1899. 


VOLUME   OCCUPIED   BY   PRECIPITATES.  85 

Thorpe's  Dictionary  of  Applied  Chemistry  gives    the 
following  variations  of  sulphuric  acid 

o°C  10°  C  20°  C  30°  C  40°  C  50°  C 
1.857  !-846  J-835  1.825  1.816  i. 806 
In  many  cases  the  correction  is  given  for  i  °  C  over 
a  range  of  temperatures :  these  are  added  to  the  specific 
gravity  at  the  temperature  of  the  experiment  to  obtain 
the  gravity  at  4°  C.* 

VOLUME    OCCUPIED    BY    PRECIPITATES. 

In  analyses  such  as  that  of  pig  lead  for  bismuth, 
the  question  of  the  volume  occupied  by  a  precipitate  be- 
comes of  importance. 

Suppose  seventy-five  grams  of  impure  lead  is  dis- 
solved in  nitric  acid,  the  lead  precipitated  by  sulphuric 
acid,  diluted  to  one  liter  and  thoroughly  mixed.  To 
save  time  in  filtering  and  washing,  it  is  desired  to  draw 
off  for  the  determination  of  bismuth,  such  a  volume  of 
the  solution  as  shall  represent  fifty  grams  of  the  original 
sample. 

Two- thirds  of  (1000  c.c.  minus  the  volume  of  the  pre- 
cipitate) will  be  the  volume  required.  The  amount  of  im- 
purities is  small  and  the  solubility  of  lead  sulphate  will 
not  introduce  an  appreciable  error,  so  we  assume  that 
the  weight  of  lead  sulphate  will  be  proportional  to  the 
seventy- five  grams  of  pig  lead. 

75  :  x  :  :  206.92  :  302.99.     x  =  109.82  grams 
The  specific  gravity  of  PbSO4  is  6.2  therefore  the  volume 

occupied  by  the  precipitate  is  — ^ or  17.7  c.c.     So  to 

*  For  a  table  of  corrections  for  alcohol  see  Zeit.  fUr  Analytische  Chemie,  vol. 
C8,  p.  253,  1899. 


86  VOLUME   OCCUPIED   BY  PRECIPITATES. 

get  a  solution  containing  the  bismuth  from  fifty  grams  of 
pig  lead,  we  must  draw  off  or  filter  through  a  dry  filter 
2/3  (ioao  —  17.7)  or  654.86  c.c. 

If  we  have  a  solution  containing  a  precipitate  and 
wish  to  titrate  the  excess  of  precipitant  in  a  portion  of 
the  solution,  we  dilute  to  the  mark  in  a  flask,  mix,  and 
then  draw  off  a  portion  wiih  a  pipette.  Suppose  the 
solution  is  diluted  to  200  c.c.  and  50  c.c.  withdrawn; 
this  is  evidently  more  than  one  quarter  on  account  of 
the  volume  occupied  by  the  precipitate.  This  error  is 
usually  disregarded,  but  it  can  be  found  as  follows.  If  we 
have  precipitated  the  lead  from  one  gram  of  lead  ore  as 
PbCrO4,  by  a  known  amount  of  K2Cr2O7,  and  wish  to 
determine  the  excess  by  titration  with  potassium  iodide 
and  sodium  thiosulphate  in  a  portion  of  the  filtrate;  in- 
stead of  filtering,  washing  and  then  diluting  to  a  known 
volume  we  may  proceed  as  follows :  after  adding 
K2Cr2Or  in  excess,  dilute  to  200  c.c.  mix  thoroughly  and 
allow  the  precipitate  to  settle;  then  withdraw  50  c.c.  of 
the  clear  solution  and  titrate  the  excess  with  sodium  thio- 
sulphate after  adding  potassium  iodide  and  hydrochloric 
acid.  This  shows,  we  will  assume,  an  excess  of  i  oo  mgs.  of 
K2Cr2O7 ;  on  the  assumption  that  we  have  titrated  one 
quarter  of  the  solution,  the  total  excess  of  K2Cr2Or 
will  be  400  mgs.  This  is  subtracted  from  the  known 
weight  of  K2Cr2O7  added  and  the  lead  present  calcu- 
lated from  the  remainder  in  the  usual  way,  giving  say 
80%  of  lead. 

Now  that  we  have  determined  approximately  the 
lead  present,  we  can  from  this  calculate  the  volume  occu- 
pied by  the  lead  chromate  and  so  correct  this  percentage 
of  lead  for  the  volume  occupied  by  the  precipitate.  80% 
Pb  =  0.8  gram  Pb  in  one  gram  =  1.250  grams  PbCrO4. 


VOLUME  OCCUPIED  BY  PRECIPITATES.  87 

The  specific  gravity  of  PbCrO4  is  6,  hence  the  volume 
occupied  by  the  precipitate  is  -— ^  or  0.21  c.c.  50  c.c. 

showed  an  excess  of  100  mgs.  of  K2Cr2Q7,  to  get  the  cor- 
rect total  excess,  we  have —  50  :  200  —  0.21  :  :  100  :  x. 
x  =  399.58  mgs.  or  0.42  milligram  more  K2Cr2O7  was 
used  in  precipitating  PbCrO4  than  was  found  before. 
Which  is  equivalent  to  0.59  milligram  of  lead,  so  that 
the  correct  weight  of  lead  present  was  0.80059  gram> 
equivalent  to  80.059%. 

It  will  be  observed  that  this  correction  depends 
mainly  on  two  things — the  excess  of  reagent  and  the 
dilution.  If  the  excess  is  small  and  the  dilution  consid- 
erable, the  error  due  to  the  volume  of  the  precipitate  can 
be  neglected. 

The  accuracy  of  this  method  of  calculating  the 
volume  of  the  precipitate  is  questioned  by  Ostwald,  on 
account  of  the  variation  of  the  specific  gravity  of  the 
precipitate  in  the  solution,  due  mainly  to  adsorption. 
The  method  however  is  much  used  in  commercial  analy- 
tical work  with  satisfactory  results. 


EXAMPLES. 


1.  A  piece   of  barite  (heavy  spar)  weighs  in  air  47 
grams  and  in  water  36.555  grams.     What  is  its  specific 
gravity?  Ans.     4.5. 

2.  A  specimen  of  magnetite  weighed  in  air  27  grams 
and  in  water  21.60  grams.     What  is  its  specific  gravity? 

Ans.      5. 

3.  A  piece  of  loaf  sugar  weighed   in   air  7.5  grams 
and   in  petroleum  ether,  Sp.  Gr.  0.645,   4.4635  grams. 
What  is  its  specific  gravity  ?  Ans.      1.593. 

4.  A   sample  of  magnetic  sand  weighed  in   air   12 
grams.     A  flask  filled  with  water  to  a  mark  weighed 
47.5327   grams.     The  same  flask  containing  the  sand 
and  filled  to  the  same  mark  with  water  weighed  56.4281 
grams.     What  is  the  specific  gravity  of  the  sand  ? 

Ans.     3.864. 

5.  To  find  the  specific  gravity  of  a  piece  of  freshly 
cut  pine  wood.     Given  : 

Weight  in  air,  ....          10.100  grams 

Weight  of  sinker  in  water,  .          .  12.24  grams 

Weight  of  wood  and  sinker  in  water,  11.217  grams 

Ans.     0.908. 

6.  To  find  the  specific  gravity  of  common  salt,  given 
the  following  data : 

Weight  of  salt  in  air,  .         .         .         8.8450  grams 

Weight  of  flask  and  turpentine  to  mark,    45.2375  grams 

(88) 


EXAMPLES.  89 

Wt.  of  flask,  salt  and  turpentine  to  mark,  50.5609  grams 
Specific  gravity  of  turpentine,  0.86 

Ans.      2.16 

7.  To  find   the  specific  gravity  of  metallic  sodium. 
Given  : 

Weight  in  dry  air,  .          .          .          14. 15629  grams 

Weight  in  kerosene,  .  .          .          2. 1807  grams 

Specific  gravity  of  kerosene,  .  .          .          0.829 

Ans.     0.98. 

8.  To  find  the  specific  gravity  of  turpentine.     Given: 
Weight  of  pyknometer,     .          .         ..          25.0456  grams 
Wt.  of  pyknometer  filled  with  water,     .     72.1485  grams 
Wt.  of  pyknometer  filled  with  turpentine,  65.5541  grams 

Ans.     0.86. 

9.  A  piece  of  silver  weighed  47  grams  in  water  and 
48.756  grams  in  petroleum  ether,  specific  gravity  0.645. 
What  is  the  specific  gravity  of  the  silver  ? 

Ans.      10.5. 

10.  A  ten  gram  brass  weight,  Sp.  Gr.  8.5,  weighed 
in  glycerin  8.5135  grams.     What  is  the  specific  gravity 
of  glycerin?  Ans.      1.2635. 

11.  Given  the  specific  gravity  of  sulphuric   acid  as 
1.84  compared   to  water  at   15°  C.     What  is  its  specific 
gravity  compared  to  water  at  4°  C  ?     (See  tables.) 

Ans.      1.8384. 

12.  Given  the   specific  gravity  of  alcohol  at  20°  C 
referred  to  water  at  4°  C  as  0.8332.     What  is  its  specific 
gravity  referred  to  water  at  20°  C  ?      Ans.     0.83467. 

13.  How  much    phosphoric  acid,    1.7  Sp.   Gr.   and 
how  much  water  are  required  to  make  400  c.c.  of  1.18 
Sp.  Gr.  ? 

Ans.      102.85  c.c.  phosphoric  acid  and  297.15  c.c.  of 
water. 


90  EXAMPLES. 

14.  How  much   nitric  acid  1.47   Sp.    Gr.  and  water 
must  be  mixed  to  give  three  liters  of  1.08  Sp.  Gr.  ? 

Ans.      510.6  c.c.  nitric  acid;     2489.4  c.c.  water. 

15.  How  much  ammonia  water  0.9  Sp.  Gr.  must  be 
taken  to  make  with  water  four  liters  0.96  Sp.  Gr.  ? 

Ans.      1.6  liters. 

16.  We  have  two  liters  of  dilute  sulphuric  acid  1.08 
Sp.  Gr.,  which  we  wish  to  use   up  in  diluting  concen- 
trated sulphuric  acid   1.82   to  1.20  Sp.  Gr.      How  much 
concentrated  acid  is  to  be  added?       Ans.     387.1  c.c. 

1 7  What  is  the  volume  occupied   by    47  grams  of 
PbSO4.     Sp.  Gr.  6.2  ?  Ans.     7.5  c.c. 

1 8.  How  many  c.c.  of  sulphuric  acid,  1.824  Sp.  Gr. 
compared  to  water  at  15°  C,  weigh  100  grams? 

Ans.     54. 88  c.c. 

19.  Wrhat  is  the  volume  occupied  by  the  barium  sul- 
phate from  one  gram  of  FeS2?     Sp.  Gr.  BaSO4  4.5. 

Ans.     0.86  c.c. 

20.  To  two  grams  of  a  mineral  containing  barium, 
200  mgs.    of  K2Cr2O7  were  added   to  precipitate   the 
barium  and  the  solution  diluted  to  500  c.c.,  mixed  and 
allowed  to  settle  :    50  c.c.  of  the  clear  solution  was  titrated 
for  the  excess  of  K2Cr2O7  by  KI  and  Na2S2O3  showing 
4.2  mgs.     What  is   the  corrected  percentage  of  barium 
oxide?     Given  the  Sp.  Gr.  of  BaCrO4  as  3.9. 

Ans.      8.2318%.      Showing    the   correction   to    be 
insignificant. 


CHAPTER    VIII. 

CALCULATIONS   OF   GASES. 

Before  proceeding  to  any  of  the  calculations  of  gases 
or  gas  analysis  a  knowledge  of  the  following  laws  is 
necessary. 

Law  of  Boyle  or  Mariotte. 

1.  The  temperature  remaining  the  same,  the  volume 
of  a  given  quantity  of  gas  is  inversely  as  the  pressure  it 
bears. 

Law  of  Charles  or  Gay-Lussac. 

2.  The  coefficient  of  expansion  of  all  gases  is  0.00366 

-or  — !     of  the  volume  at  o°  C. 

273 

Differently  expressed,  the  volume  of  all  gases  is  di- 
rectly proportional  to  the  temperature  calculated  from 
the  absolute  zero  (-  273°  C). 

Law  of  Avogadro. 

3.  Under  the  same  temperature  and  pressure  equal 
•volumes  of  all  gases  contain  the  same  number  of  mole- 
cules. 

It  is  evident  from  (i)  and  (2)  that  the  volume  of 
gases  varies  greatly  under  different  conditions  of  tem- 
perature and  pressure  so  that  these  must  always  be  re- 
duced to  the  standard  conditions  of  o°  C  and  760  milli- 
meters pressure  for  comparison. 

(01) 


$2  CALCULATIONS   OF   GASES. 

This  is  done  as  follows :  Suppose  we  have  six  liters 
of  a  gas  at  18°  C.  and  772  m.m.  pressure  and  we  wish 
to  know  the  volume  under  the  standard  conditions.  We 
can  first  correct  for  temperature  and  then  for  pressure  or 
vice  versa.  The  volume  varies  as  the  temperature  calcu- 
lated from  the  absolute  zero,  therefore 

273  :  273  +  18  :  :  x  :  6.      x  =  5.6288. 
The  volume  at  o°  C  and  772  m.m.  is  5.6288  liters.    The. 
volume  varies  inversely  as  the  pressure,  therefore : 
772  :  760  :  :  y  :  5.6288.      y  =  5.7176  liters, 
the  volume  of  the  gas  at  o°  C  and  760  m.m.  pressure. 

These  calculations  are  necessary  in  gas  analysis  and 
in  many  other  analytical  determinations  where  the 
volume  of  a  gas  is  measured.  When  the  gas  is  measured 
over  water  it  contains  aqueous  vapor  which  supports 
part  of  the  pressure :  this  tension  increases  with  the 
temperature  (see  table),  therefore  before  proceeding  to 
the  corrections  for  temperature  and  pressure  as  before, 
the  tension  of  aqueous  vapor  must  be  subtracted 
from  the  barometer  reading  to  get  the  real  pressure 
supported  by  the  gas.  Take  the  following  example  of 
a  nitrogen  combustion.  Weight  substance  o.  1923  gram  : 
volume  nitrogen  24.6  c.c.  :  temperature  25°  C.  :  baro- 
metric pressure  771  m.m.  The  tension  of  aqueous  vapor 
at  25°  C  is  23.5  m.m.;  so  771  —  23.5  or  747.5  m.m.  is  the 
pressure  supported  by  the  nitrogen.  As  the  volume 
varies  inversely  as  the  pressure,  we  have : 

24.6  :  x  :  :  760  :  747.5.     x  =  23.93  c.c. 

(volume  at  normal  pressure} 

As  volume  varies  directly  as  the  temperature  calculated, 
from  the  absolute  zero. 

23.93  •  Y  •  ^  273  +  25  :  273.     y  =  21.95  c.c. 


DENSITY   OF   GASES.  93 

volume  of  nitrogen  at  standard  conditions,      i  c.c.  nitro- 
gen at  760  m.m.  and  o°  C  weighs  0.001254  gram  (see 
table). 
Wt.  of  nitrogen   is  o.ooi254X  21.95   =  0.02751   gram 

.and   --  --'*?  X  100  =  14.31%  nitrogen. 
0.1923 

Suppose  we  have  200  c.c.  of  gas  at  10°  C  and  765 
m.m.  and  wish  to  know  the  volume  at  18°  C  and  772 
m.m.  :  to  correct  for  temperature,  we  .have: 

200  :  x  :  :  273  +  10  :  273  -f-  18.     x  =  202.08  c.c. 
and  lor  pressure 

202.08  :  y  :  :  772  :  765.      y  =  200.24  c.c. 
the  volume  under  the  desired  conditions. 

DENSITY    OF    GASES. 

The  density  or  specific  gravity  of  gases  is  referred 
sometimes  to  air  at  760  m.m.  pressure  and  o°  C  as 
unity  and  sometimes  to  hydrogen  under  the  same  con- 
ditions. In  the  actual  determinations  it  is  usually  com- 
pared with  an  equal  volume  of  air  and  then  referred  to 
hydrogen  by  calculation.  Since  a  liter  of  hydrogen 
weighs  0.08952  gram  under  the  standard  conditions  and 
a  liter  of  air  1.29305  grams,  the  density  referred  to  air 
can  be  referred  to  hydrogen  by  multiplying  by  14.44  * 

The  important  relation  between  the  molecular  weight 
and  density  of  gases  referred  to  hydrogen  has  already 
been  mentioned  under  the  calculation  of  atomic  weights, 
as  a  means  of  determining  what  multiple  of  the  equivalent 
is  the  atomic  weight ;  it  is  also  one  of  the  most  import- 
ant methods  for  determining  molecular  weight.  When 

••  This  factor  is  variously  given  from  14.438  to  14.448. 


94  DENSITY   OF   GASES. 

we  say  the  density  of  a  gas  is  40  referred  to  hydrogen, 
it  means  that  any  given  volume  of  the  gas  is  40  times  as 
heavy  as  the  same  volume  of  hydrogen,  under  the  same 
conditions  of  temperature  and  pressure.  Now  accord- 
ing to  the  law  of  Avogadro,  these  equal  volumes  of 
gases  contain  the  same  number  of  molecules  so  each 
molecule  of  the  gas  is  40  times  as  heavy  as  a  molecule 
of  hydrogen,  and  as  the  hydrogen  molecule  contains  two 
atoms,  a  molecule  of  the  gas  is  80  times  as  heavy  as  a 
hydrogen  atom  or  its  molecular  weight  is  80.  Or,  in 
brief,  the  density  of  a  gas  (referred  to  hydrogen)  is  one 
half  its  molecular  weight. 

The  exact  calculation  of  density  and  so  molecular 
weight  varies  with  the  method  used,  but  in  general  it  is 
as  follows : 

We  have  given  the  weight  of  a  known  volume  of  gas 
under  certain  conditions  of  temperature  and  pressure. 
Reduce  these  to  o°C  and  760  m.  m.,  and  divide  the 
weight  by  that  of  an  equal  volume  of  hydrogen  and  we 
have  the  density  or  specific  gravity.  The  density  of 
vapors  is  determined  in  a  similar  way,  except  that  the 
volume  occupied  by  the  vapor  from  a  known  weight  of 
substance,  is  measured  instead  of  the  weight  of  a  known 
volume  of  vapor  or  gas.  In  many  cases,  especially  when 
the  heat  necessary  to  convert  the  substance  into  the 
gaseous  condition  is  high,  it  is  more  convenient  to  meas- 
ure the  volume  of  air  displaced  by  the  substance  in  the 
gaseous  form,  rather  than  the  gas  itself;  but  the  calcu- 
lation is  the  same,  for  the  volume  occupied  by  the  air  is 
the  same  as  would  be  occupied  by  the  vapor  from  the 
known  weight  of  the  substance  at  the  same  temperature 
and  pressure. 

The  following  determinations  of  the  density  of  chloro- 


DENSITY   OF   GASES.  95 

form  vapor  by  the  three  principal  methods,  illustrate 
these  calculations. 

(i)   Dumas'  method.      Given  the  following  data. — 

Weight  of  bulb  filled  with  air  at  i9°C  and  763.9 
m.  m.  =  36.4489  grams. 

Weight  of  bulb  filled  with  vapor  at  ioi°C  and  763.9 
m.  m.  =  37.2085  grams. 

Volume  of  air  in  the  bulb  at  I9°C  =  279  c.c. 

First  find  the  weight  of  the  air  contained  in  the 
bulb  ;  by  reducing  the  volume  to  the  standard  conditions 
and  multiplying  by  0.001293,  the  weight  of  i  c.c.  of  air 
at  o°C  and  760  in.  m. 

279  :  x  :  :  273  -f-  19  :      273.      x  =  260.8  c.c. 
260.8  :  y  :  :        760          :   763.9.      y  =  262.1  c.c. 

Weight  of  air  is  262.1  X  0.001293  =  0.3389  gram. 

Weight  of  bulb  is  36.4489  —  0.3389  =  36.1100  grams. 

Weight  of  vapor  is  3  7. 2085  -  36. 1 1  oo  =  i  .0985  grams 

We  now  know  that  279  c.c.  of  chloroform  vapor  at 
ioi°C  and  763.9  m.  m.  weighed  1.0985  grams. 

To  find  the  corresponding  volume  under  the  stand- 
ard conditions,  we  have  : 

279  :  x  :  :  273  +  101  :  273.      x  =  203.8  c.c. 
203.8  :  y  :  :  760  :  763.9.      y  =  204.8  c.c. 

The  weight  of  an  equal  volume  of  air  will  be  204.8 
X  0.001293  or  0.2648  gram. 

So  1.0985  -r-  0.2648  or  4.148  is  the  density  referred 
to  air:  4.148  X  14.44  or  59.897  is  the  density  referred  to 
hydrogen  and  59.897X2  or  119.794  is  the  molecular 
weight.  Theory  119.37. 

(2.)  Victor  Meyer's  method.  Given  the  following 
data. 


96  DENSITY   OF   GASES. 

Weight  of  chloroform  taken  0.2097  gram. 

Volume  of  air  expelled  by  vapor  43.65  c.c. 

Temperature  of  air  expelled  20° C. 

Height  of  barometer  763.9  m.  m. 

Tension  of  aqueous  vapor  at  2O°C  (from  table)  17.36 
m.  m. 

We  have  the  volume  of  air  expelled  by  and  equal  to 
the  volume  of  the  vapor;  to  find  its  weight  reduce  this 
volume  to  the  standard  conditions  ; 

43.65  :  x  :  :  273  +  20  :  273.      x  =  40.67  c.c. 
40.67  :  y  :  :  760  :  763.9  -  17.36.      y  =  38.50  c.c. 

38-5oX  0.001293  =  0.04978  gram:  weight  of  air. 

Then  0.2097  -f-  0.04978  is  4.21,  the  density  referred 
to  air  ; 

4.21  X  14-44  =  60.79,  tne  density  referred  to, 
hydrogen ; 

60  79  X  2  =  121.58,  the  molecular  weight. 

(3.)  Hofmann's  method.  The  following  data  are  re- 
quired : 

Weight  of  chloroform  o.  r  535  gram. 

Volume  of  vapor  at  ioo°C  79  c.c. 

Height  of  barometer  at  2i°C  774.8  m.  m. 

Height  of  column  of  mercury  (inside  heating 
jacket)  at  ioo°C  246.5  m.  m. 

Height  of  column  of  mercury  (outside  heating 
jacket)  at  25°C  143.8  m.  m. 

*  Tension  of  mercury  vapor  at  ioo°C  0.27  m.  m. 

This  calculation  involves,  besides  those  already  dis- 
cussed, the  reduction  of  the  barometric  readings  to  o°C  ; 
because  they  are  at  such  widely  varying  temperatures. 
The  coefficient  of  the  expansion  of  mercury  is  0.00018 


Ostwald  I'hysico-chemical  measurements  p.  100. 


DENSITY   BY   EFFUSION.  97 

Tor    i°C  :   so    to    reduce   the    reading   representing    the 
atmospheric  pressure  to  zero,  we  have  : 

i  :  i  -J-  (21  X  0.00018)  :  :  x  :  774.8  or 

774.8  -4-  1.00378  =  771.88  m.  m. 
For  the  other  columns  of  mercury  we  have  : 
246.5  -f-  1.018  =  242.14  m.    m.   and    143.8  -5-  1.0045  = 

143.15  m.  m. 

The  79  c.c.  of  vapor  is  under  the  pressure  indicated 
by  the  barometer  minus  the  tension  of  mercury  vapor 
(0.27  m.  m.)  and  minus  the  sum  of  the  two  columns  of 
mercury  also  supported  by  the  atmospheric  pressure 
(242.14-)-  143.15).  So  the  pressure  is  771.88  —  (0.27 
+242.14+143.15)  =  385-56  m-  m- 

We  now  know  that.  0.1535  gram  of  vapor  at  ioo°C 
and  385.56  m.  m.  occupy  79  c.c.  Reduce  this  as  before 
to  o°C  and  760  m.  m.  pressure  and  divide  the  weight 
by  that  of  an  equal  volume  of  air  : 

79  :  x  :  :  373  :  273.     x  =  57.82  c  c. 
57.82  :  y  :  :  760:  385.56.     y  =  29.333  c.c. 
29-333  X  0.001293  =  0.037927 
ai535  ^~  o-O37927  =  4-°47  the  density  re- 
ferred to  air. 

DENSITY   BY    EFFUSION. 

The  density  or  specific  gravity  of  gases  can  be  deter- 
mined with  sufficient  accuracy  for  many  practical  pur- 
poses by  the  effusion  test* 

This  is  based  on  the  fact  that  the  specific  gravity  of 
gases  is  proportional  to  the  squares  of  the  times  of  effu- 
sion. In  these  tests  it  is  most  convenient  to  determine 

*  For  details  see  Hempcl's  Gas  Analysis,  Dennis  p.  212. 


98  CORRECTION    OF   WEIGHINGS. 

the  specific  gravity  with  reference  to  air  and  then,  if 
necessary,  refer  it  to  hydrogen  by  calculation. 

Suppose  a  certain  volume  of  air  requires  twenty 
minutes  to  escape  through  an  opening  made  by  a  needle 
in  a  platinum  foil,  and  that  under  the  same  conditions  of 
temperature  and  pressure  the  same  volume  of  another 
gas  requires  twelve  minutes. 

Let  x  =  density  of  this  gas  :  then 

x  :  i  :  :  144  :  400.  x  =  0.36,  density  referred  to 
air,  and  0.36  X  H-44  =  5-198,  density  referred  to 
hydrogen. 

This  test  is  much  used  for  illuminating  gas  at  the 
works,  but  is  not  sufficiently  accurate  for  scientific 
work. 

CORRECTION    OF   WEIGHINGS. 

According  to  the  law  of  Archimedes,  when  any 
object  is  weighed  under  the  ordinary  conditions  in  air, 
we  obtain  less  than  the  true  weight  on  account  of  the 
loss  in  weight  of  the  object  equal  to  that  of  the  air  dis- 
placed. This  correction  is  small,  but  it  must  be  made 
in  some  cases,  especially  when  large  light  apparatus  is 
weighed,  such  as  absorption  bulbs  &c.,  and  in  all  very 
accurate  determinations,  such  as  atomic  weights. 

The  correction  is  therefore  the  addition  of  the  weight 
of  an  equal  volume  of  air  at  the  temperature  of  the 
weighing.  This  may  be  done  as  follows :  weight  in  air 
then  in  water  ;  the  loss  in  weight  is  equal  to  the  weight 
of  the  same  volume  of  water;  now,  at  the  ordinary 
laboratory  temperature  the  density  of  air  referred  to 
water  as  unity  is  0.0012,  so  the  loss  in  weight  of  the 
substance  in  water  times  o.ooi  2*  is  the  correction  to  add. 


*  This  weight  of  1  c.c.  allows  for  mean  humidity  as  well  as  temperature. 


CORRECTION   OF   WEIGHINGS.  99 

If  the  specific  gravity  of  the  object  is  known  the  cor- 
rection can  be  calculated  as  follows  : 

Wt.  in  air  *  , 

_  =  volume  in  c.c. 
Sp.  Gr. 

Volume  times  weight  of  i  c.c.  of  air  (0.0012)  is  the 
correction  to  add. 

In  making  corrections  for  weighing  it  must  be  borne 
in  mind  that  we  have  : 

(a)  The  apparent  weight,  that  actually  found. 

(b)  The   true  weight  in  vacuo,  obtained  by   calcu- 
lation. 

To  get  the  true  weight  of  any  object,  we  must  add  to 
its  apparent  weight  the  weight  of  an  equal  volume  of  air 
and  subtract  the  weight  of  a  volume  of  air  equal  to  that 
of  the  weights  used.  The  correction  for  the  weights 
diminishes  the  correction  for  substances  having  a  lower 
specific  gravity  than  that  of  the  weights;  when  the  den- 
sities are  the  same  the  corrections  equalize  each  other  : 
when  the  density  of  the  object  is  greater  than  that  of 
the  weights  the  correction  becomes  negative  (is  to  be 
subtracted). 

Suppose  we  have  a  100  gram  brass  weight  correct 
in  air  ;  it  will  weigh  slightly  more  in  vacuo.  To  find 
this  weight  we  have  : 

rr  •   i  fc  •   i       i     (true  wt.  x  0.0012) 

1  rue  weight  =  apparent  weight  -j-  v  -  .  ---- 

Sp.  Gr. 

=  0.0141  ;   so  that  the   true 


weight  in  vacuo  is  100.0141  grams. 

Let  us  use  this  weight  to  weigh  different  substances 
and  calculate  their  true  weight. 


*  The  method  given  is  all  that  is  ever  required  but  is  not  strictly  accurate, 
for  the  true  weight  should  be  used  instead  of  the  weight  in  air. 


100  CORRECTION   OF   WEIGHINGS. 

I st.    Sulphur,  specific  gravity  2.      The  correction  will 

T  S~\f\ 

be  volume  X  0.0012    or  —   X  0.0012  =  0.06  so  that  the 

2 

true  weight  becomes 

100  +  (0.06  —  0.0141)  =  100.0459. 

We  see  from  this  that  the  true  weight  is  that  of  the 
brass  weight  in  air  plus  the  weight  of  air  displaced  by  the 
object  less  that  displaced  by  the  weight  itself. 

2nd.  If  we  balance  the  100  gram  brass  weight  with 
brass  of  the  same  specific  gravity,  we  shall  have  the 

correction  to  add  -     -  X  0.0012  =  0.014  and  that  to  de- 

8-5 

duct  0.014,  so   that   the  true  weight  coincides  with  the 
apparent  weight. 

3rd.  If  we  balance  the  100  gram  brass  weight  with 
platinum,  specific  gravity  21.5,  we  shall  find  the  true 
weight  less  than  100  grams  ;  for  the  correction  to  add  is 

X  0.0012  =  0.00^6;    while  that  to  subtract  is  as  be- 

21-5 

fore  0.014  :    hence  the  true  weight  becomes  100  -f-  0.0056 

-  0.014  =  99.9916  grams. 

It  will  be  seen  from  the  preceding  that  the  correction 
is  different  for  brass  and  for  platinum  weights,  and  that 
the  correction  becomes  greater  as  the  specific  gravity 
become  less. 

Suppose  we  have  a  glass  vessel  whose  apparent 
weight  is  75  grams  when  weighed  with  brass  weights, 
and  whose  specific  gravity  is  2.5.  What  is  the  true 
weight  ? 

The  volume  is  2*.  or  30.  c.c.  ;  the  weight  of  an  equal 
volume  of  air  is  30  X  0.0012  =  0.036  gram:  the  correc- 
tion for  the  brass  weights  is  ^1  X  0.0012  =  0.0105 


CALCULATIONS   OF    GAS    ANALYSIS. 


101 


gram  :   so  the  true  weight  becomes  75  +  (0.036  —0.0105) 
=  75-0255  grams. 

The  following  table  gives  the  weight  in  milligrams 
to  be  added  or  subtracted  for  each  gram  of  apparent 
weight  of  substance,  for  different  specific  gravities,  when 
weighed  in  air  with  brass  weights.* 


Sp.  Gr. 

Correction. 

Sp.  Gr. 

Correction. 

Sp.  Gr. 

Correction. 

0.7 

+  1.57 

2.0 

+   0.458 

9. 

—  0.009 

0.8 

+  1.36 

2".  5 

-f   0.337 

10. 

—  0.023 

0.9 

+  1.19 

3.0 

+   0.257 

11. 

-  0.034 

1.0 

-j-  1.06 

3.5 

4-   0.200 

12. 

-  0.043 

1.1 

-j-  0.95 

4.0 

4-   0.157 

13. 

-  0.050 

1.8 

-J-  0.86 

4.5 

4-   0.142 

14. 

—  0.057 

1.8 

-f  0.78 

5.0 

+    0.097 

15. 

-   0.063 

1.4 

-h  0.71 

5.5 

4-   0.075 

16. 

—  0.068 

1.5 

-j-  0.66 

6.0 

4-    0.057 

17. 

-  0.072 

1.8 

4-  0.61 

6.5 

4-   0.042 

18. 

—  0.076 

1.7 

-f  0.56 

7.0 

-f   0.02!) 

19. 

—  0.080 

1.8 

-j-  0.52 

7.5 

4-   0.017 

20.            -   0.083 

1.9 

+  0.49 

8.0 

-j-   0.007 

CALCULATIONS     OF    CAS    ANALYSIS. 

With  the  help  of  the  law  of  Avogadro  we  can  very 
readily  understand  the  combination  of  gases  by  volume. 
If  we  cause  two  liters  of  carbonic  oxide  to  combine  with 
one  liter  of  oxygen,  the  product  is  not  three  liters  of 
carbonic  acid  but  two ;  for  there  are  present  the  same 
number  of  molecules  of  CO2  as  there  were  of  CO  and  as 
equal  volumes  of  all  gases  contain  the  same  number  of 
molecules,  an  equal  number  of  molecules  occupy  the 


*  Calculated  by  Kohlrausch.     Ostwald's  Physico-Chemical  Measurements. 


102  CALCULATIONS   OF   GAS   ANALYSIS. 

same  volume,  at  the  same  temperature  and  pressure : 
hence  we  have 

2  vols.  CO  -f  i  vol.  O,  =  2  vols.  COo. 
This  shows  that  the  combination  of  the  two  gases  is  at- 
tended by  a  decrease  in  volume  and  that  the  resulting 
volume  is  two -thirds  of  that  occupied  by  the  mixture. 

Similarly  we  have 

2  vols.  Ho  -j-  i  vol.  (X  =  2  vols.  H2O  (as  steam) 

i  vol.  CH4  +  2  vols.  6,,  =  i  vol.  CO.+  2  vols.  HoO 
(as  steam)  or  if  the  steam  be  allowed  to  condense 

i  vol.  CH4  +  2  vols.  O._,  =  i  vol.  CCX  -|-  water. 

This  law  together  with  the  corrections  for  tempera- 
ture, pressure  and  tension  of  vapor,  already  explained, 
furnish  all  the  information  necessary  to  calculate  gas 
analyses. 

For  example  let  us  calculate  the  percentages  from 
the  following  analytical  results  on  a  sample  of  water 

gas- 
Volume  taken  for  analysis,          .  .         .          100  c.c. 
Contraction  due  to  absorption  of  C(X,          .  o.o  c.c. 
"           "  illuminants,       12.6  c.c. 

"     "  "  oxygen,      .          0.9  c.c. 

"    "  "  CO,  .        27.3  c.c. 

We  have  left  59.2  c.c.  of  the  original  100  c.c.  containing 
hydrogen,  methane  &c.,  and  nitrogen,  16.2  cc.  of  this 
are  withdrawn  to  the  explosion  tube  and  mixed  with 

32.3  c.c.  of  oxygen  and  then  diluted  with  31.4  c.c.  of  air 
giving  a  total  volume  of  80.  i  c.c.  :    this  is  then  exploded 
and  after  allowing  to  cool   the  volume  is  found  to  be 

53.4  c.c. 

We  h-ave  in  the  tube  before  explosion. 
x  Vols.  CH4  -f  y  vols.  H2  -f-  m  vols.  N,  +  n  vols.  O,. 


CALCULATIONS   OF   GAS   ANAYLSIS.  103 

After  explosion  this  becomes 

x  vols.   CO3  +  (2  x  4-  y)  vols.  H2O  (condensed)  +  m 

vols.  No  (unaffected)  -f  n   -  (2  x  +  y2  y)  vols,  O2. 

The  volume  of  oxygen  is  diminished  by  a  volume 
equal  to  twice  that  of  the  CH4  plus  one-half  that  of  the 
hydrogen. 

The  CO.,  produced  by  the  explosion  is  removed  by 
dilute  potassium  hydroxide  solution  and  the  volume  of 
the  gas  again  measured  =  45.8  c.c.  The  contraction  in 
volume  produced  by  the  removal  of  C(X  equals  the 
volume  of  CH4  present,  7.6  c.c.  The  contraction  in 
volume  due  to  the  explosion,  26.7  c.c.,  therefore  repre- 
sents the  volume  occupied  by  the  hydrogen  (condensed 
to  water)  and  the  decrease  of  oxygen  equal  to  twice  the 
volume  of  CH4  plus  one  half  the  volume  of  hydrogen. 
The  CO._>  produced  takes  the  place  of  the  CH4  originally 
present. 

If  we  let  x  =  vol.  of  hydrogen,  we  have  : 

26.7  =  x  +  2(7.6)  +  YZ  x-      x  =  7-66  c.c. 
Or,  the  volume  of  hydrogen  =  %  of  contraction  minus 
twice  the  volume  of  CO.,  formed  (second  absorption  by 
KOH). 

As  only  a  portion  of  the  residual  gas  was  taken  for 
the  explosion,  the  volumes  obtained  are  referred  to  the 
original  100  c.c.  by  proportions: 

for  CH4,   1 6. 2  :  59.2  :  :  7.6    :  x.      x  =  27.77  c-c- 
for  Ha,      16.2  :  59.2  :  :  7.66:  y.     y  =  27.99  c-c- 

The  results  expressed  in  percentage  by  volume  are  : 
carbonic  acid,  none  :  illuminants,  1 2.6%  :  oxygen,  0.9% : 
carbonic  oxide,  27.3%  :  hydrogen,  28%  :  methane  &c., 
27.7%:  nitrogen  (by  difference)  3.5%. 

For  accurate  work  all  volumes  should  be  corrected 


104  CALCULATIONS   OF   GAS   ANALYSIS. 

for  temperature,  pressure,  and  vapor  tension  as  already 
explained. 

The  following  are  results  on  coal  gas  with  the  Elliott 
apparatus : 

Volume  taken  for  analysis  100  c.c. 

Reading  after  treatment  with  KOH.  (CO.,)  o.o  c.c. 

"  "  "  "      bromine  (illuminants) 

6.5  c.c. 

Reading  after  treatment  with  alkaline  pyrogallate 
(02)  7-4  c.c. 

Reading  after  treatment  with  CuoCU  and  HCli 
(CO)  14.2  c.c. 

Drawn  off  for  combustion  15.8  c.c. 
Volume  after  adding  oxygen  48.0  c.c. 
"  "          "         air  79.7  c.c. 

"  "     explosion  53.5  c.c. 

"  "     washing  with  KOH  46.0  c.c. 

The   volume  of  CH4  =  CO.,  formed  =  53.5—46  = 

7-5  c-c- 

As  the  volume  of  hydrogen  =  ^  of  contraction  — 
twice  the  CO2  formed :  we  have, 

Volume  of  hydrogen  =  %  (26.2  —  15)  =  7.46  c.c. 

To  refer  these  quantities  to  the  original  100  c.c.  we 
have,  residual  volume  =  100  -  14.2  c  c.  =  85  8  c.c.  and 

for  H2,  15.8  :  85.8  :  :  7.46  :  :  x.      x  =  40.5 
for  CH4,  15.8  :  85.8  :  :     7.5  :  :  y.      y  =  40.7 

So  our  results  become  :  carbonic  acid,  none  :  illum- 
inants, 6.5%:  oxygen,  (7.4  —  6.5)0.9%:  carbonic  oxide, 
(14.2-7.4)6.8%:  hydrogen,  40.5%:  methane  &c.,, 
40.7%;  nitrogen  (by  difference),  4.6%. 


r 


EXAMPLES. 


1.  Reduce  twelve  liters  of  oxygen,  measured  at  47°C 
and  under  a   pressure  of  782   m.m.  of  mercury,  to  the 
standard  conditions.  Ans.      10.5338  liters. 

2.  What  will  be  the  volume  of  nine  liters  of  hydro- 
gen (at  the  standard  conditions)  when  heated  to  I2O°C 
and  under  a  pressure  of  423  m.m.  ? 

Ans.     23.278  liters. 

3.  Given  27  c.c.  of  nitrogen  at  i4°C  and  752.1   m.m. 
What  will  be  its  volume  at  2o°C  and  767.8  m.m.  ? 

Ans.     27  c.c. 

4.  What  is  the  weight  of  7 7. 2   c.c.  of  air  measured 
over    water   at    27°C    and     758     m.m.    pressure?     See 
tables.  Ans.     0.08742  gram. 

5.  What   is    the    weight  of  1 7   liters  of  hydrogen, 
measured  at  i2°C  and  3  atmospheres  pressure?     Take 
i  atmosphere  =  760  m.m  of  mercury. 

Ans.     4.37342  grams. 

6.  What  is  the   weight  of  85  c.c.  of  NH3  gas  meas- 
ured over  mercury  at  i4O°C  and  under  a  pressure  of  770 
m.m.  ?     Given  the  tension  of  mercury  vapor  at  I4O°C  = 
1.76  m.m.  Ans.     0.04324  gram. 

7.  Find    the   density  referred  to   air  of  chloroform 
vapor.      Given  the  following  data  for  Dumas'  method. 

Weight  of  bulb  and   air  at  2O-5°C   and  763.9  m.m. 
34.8451  grams. 

(105) 


106  EXAMPLES. 

Weight   of  bulb  and  vapor   at    u6.5°C   and  763.9 
m.m.  3 5. 8430  grams. 

Volume  of  air  in  bulb  at  2O.5°C  394.93  c.c. 

Ans.     4.10 

8.  What  is  the  density   referred  to  hydrogen  of  a 
substance  which  gave  the  following  results  by  Victor 
Meyer's  method  ? 

Weight  of  substance  taken  0.1307  gram. 

Volume  of  airexpelled  (measured over  water)  25.8  c.c. 

Temperature  of  air  expelled  2O°C. 

Height  of  barometer  763.9  m.m.  Ans.     61.8. 

9.  What  is   the    molecular  weight  of   a    substance 
which  gave  the  following  results  by  Hofmann's  method. 

Weight  of  substance  used  0.1002  gram. 
Volume  of  vapor  62. 35  c.c. 
Height  of  barometer  at  2i.5°C  771.5  m.m. 
Height  of  column  of  mercury — (inside  heating  jacket) 
at  ioo°C  314  m.m. 

Height  of  column  of  mercury — (outside  heating  jacket) 
at  26.5°C  135  m.m. 

Tension  of  mercury  vapor  at  ioo°C  0.27  m.m. 

Ans.      114.5. 

10.  What  is   the  density   referred  to  air  of  a  gas, 
which  escapes  in    14.5    minutes    through    an    opening, 
through  which,  at  the  same  temperature  and  pressure, 
the  same  volume  of  air  requires  19.5  minutes  ? 

Ans.     0.55. 

11.  What  is   the   density  referred  to  hydrogen  of  a 
gas,  which  escapes   in    7    minutes   30   seconds  from  an 
opening,  while   the  same  volume  of  air  under  the  same 
conditions  requires  5  minutes  10  seconds? 

Ans.     30.44. 


I 


EXAMPLES.  107 

12.  An  object  whose  specific  gravity  is  3.38  (H2O  = 
i.),  weighed  39.7250  grams  in  air,  when  weighed  with 
brass  weights.     What  is  its  true  weight  ? 

Ans.     39-7335  grams. 

13.  An  object  whose  specific  gravity  is  0.95  weighed 
i7.854Ograms  in  air,  when  weighed  with  brass  weights. 
What  is  its  true  weight  ?  Ans.      17.8740  grams. 

14.  A   bar  of  gold   Sp.    Gr.  19.3  weighed  10.73685 
kilos,  when  weighed  in  air  with  brass  weights.     What 
is  the  true  weight  ?  Ans.     10.73600  Kilos. 

15.  A  piece  of  pyrrhotite,  Sp.  Gr.  4.5,  weighed  in  air 
47.3854   grams,  when   weighed  with  platinum  weights. 
What  is  its  true  weight  ?  Ans.     47.3954  grams. 

1 6.  A  piece  of  antimony,  Sp.  Gr.  6.8,    weighed  75 
grams  in  air;  the  weights  used  were  50  grams  brass,  20 
and  5  grams  platinum.     What  is  the  true  weight  ? 

Ans.     75.0048. 

17.  If  50  c.c.   of  CH4,    50  c.c.  of  H2  and  5oc.c.  of 
C2H4  were  mixed  with  450  c.c.  of  oxygen  and  exploded. 
What  would   be  the  resulting  volume  at  the  same  tem- 
perature and  pressure  ?     (Assume  that  the  temperature 
is  below  ioo°C.).  Ans.     325  c.c. 

1 8.  Calculate  the  percentage   by  volume  of  a  blast 
furnace  gas,  giving  the  following  results: 

Taken  for  analysis  100  c.c. 
Volume,  after  washing  with  KOH  (CO2)  99.4  c.c. 

"  "  "          "     Br  (illuminants)  99. 4  c.c. 

"  "  "         "  Alkaline  "pyro"  (Oo)99-4c.c. 

«Cu2Cl2,HCl(CO)65.ic.c. 

Decrease  in  volume  after  explosion 

(the  whole  used)  2.1  c.c. 

Removed  by  washing  with  KOH  none. 
Ans.     CO,  0.6%.     CO  34.3%.     H,  1.4%.     N863.7%. 


I 


108  EXAMPLES. 

19.  Calculate  the   percentage   by  volume  of  a  pro- 
ducer gas  giving  the  following  results  : 

Volume  taken  for  analysis  looc.c. 

after  washing  with  KQH  (CO2)  98.5  c.c. 

"     Br  (illuminants)  98.5  c.c. 
"    Alk.  "pyro."  (CX)  98.5  c.c. 
"   CuXl2,  HC1  (GO)   75.0  c.c. 

Decrease  in  volume  after  explosion,  the  whole  used. 
15  c.c.  Removed  by  washing  with  KOH  3  c.c. 

Ans.  CO,,  1.5%.  CO,  23.5%.  CH4,  3%.  H,,  6%. 
N2  by  difference,  66%. 

20.  Calculate  the  composition  of  a  coal  gas  given  the 
following  data : 

Volume  taken  for  analysis  100  c.c. 

"       after  washing  with  KOH  (COo)  99.5  c.c. 

"  "  "          "     Br  (illuminants)  95.5  c.c. 

"     Alk.  "pyro."  (O2)  95.0  c.c. 
"      CugCls,  HC1  (CO)  89.0  c.c. 

Decrease  after  explosion,  one  quarter  used,  36.9  c.c. 

Removed  by  washing  with  KOH  10.  i  c.c. 

Ans.  CO2,  0.5%.  Illuminants  (C2H4  &c.),  4%.  O2, 
0.5%.  CO,  6%.  CH4&c.,  40.4%.  H,,  44.5%.  No,  by 
difference,  4.1%. 


CHAPTER   IX. 

CALCULATIONS   OF   CALORIFIC    POWER. 

Before  taking  up  the  calculations  of  the  calorific 
power  of  the  usual  fuels,  a  few  words  on  thermo-chemis- 
try  are  necessary,  to  serve  as  an  introduction  to  this  and 
the  following  chapter.  We  have  already  discussed  the 
qualitative  and  quantitative  aspects  of  a  chemical  re- 
action ;  it  remains  to  show  here  the  thermo-chemical 
relations  expressed. 

All  chemical  reactions  are  accompanied  by  the  liber- 
ation or  absorption  of  a  definite  amount  of  energy,  which 
may  manifest  itself  in  various  ways,  such  as  heat  or  elec- 
tricity. Let  us  consider  the  equation  : 

Zn-|-  CL  =  ZnCL,  as  written  this  signifies  that  65.41 
parts  by  weight  of  zinc  combine  with  70.9  parts  of 
chlorine  to  give  136.31  parts  of  zinc  chloride  ;  but  be- 
sides this,  heat  is  liberated  equivalent  to  97,200  calories, 
when  the  quantities  used  are  the  atomic  or  molecular 
weights  in  grams,  so 

Zn  -j-  CL  =  ZnCL  +  97,200  calories. 
In  other  words  65. 4 1  grams  of  zinc  and  70. 9  grams  of  chlo- 
rine contain  when  uncombined  potential  energy  equiva- 
lent to  97,200  calories  more  tnan  that  of  136.3 1  grams  of 
zinc  chloride  ;  consequently  to  decompose  zinc  chloride, 
into  zinc  and  chlorine,  energy  must  be  expended,  equiva- 
lent to  that  given  up  when  combination  took  place  or 
97, 200  calories  per  gram-molecule.  This  energy  can  be 

(109) 


110  CALCULATIONS   OF   CALORIFIC    POWER. 

conveniently  supplied  in  the  form  of  an  electric  current ; 
and  the  calculation  of  the  voltage  necessary  to  decom- 
pose certain  molecules  will  be  discussed  in  the  next 
chapter. 

The  heats  of  formation  of  many  molecules  have  been 
determined  and  are  usually  expressed  in  terms  of  heat 
developed  by  the  formation  of  the  molecular  weight  in 
grams  of  the  substance  ;  so  that  the  numbers  are  directly 
comparable  with  the  molecules  taking  part  in  reactions. 
This  is  not  always  the  case  however  and  care  must  be 
taken  to  ascertain  whether  the  table  is  based  on  gram- 
molecules,  gram-equivalents  or  grams  of  the  compound 
formed. 

The  unit  of  heat  is  the  calorie,  the  amount  of  heat 
necessary  to  raise  one  gram  of  water  from  4°  to  5°  C.  : 
as  the  specific  heat  of  water  does  not  vary  greatly  with 
the  temperature,  a  calorie  is  taken  practically  as  the 
amount  of  heat  necessary  to  raise  one  gram  of  water  one 
degree  centigrade.  Other  units  in  use  are :  the  greater 
calorie,  the  heat  necessary  to  raise  one  kilogram  of  water 
i°  C,  equal  to  1000  calories:  and  a  value  denoted  by 
K,  introduced  by  Ostwald,  the  amount  of  heat  necessary 
to  raise  one  gram  of  water  from  o°  to  100°  C,  this  is 
practically  equal  to  100  calories.* 

The  following  are  important  thermo-chemical  re- 
actions : 

H,  +  O  =  H,,O  +  68400  calories  (684  K) 
C  +  O  =  CO  +  29000  "  (290  K) 
CO  +  O  =  CO,  +  68000  "  (680  K) 
C  +  O,  =  CO,  -f  97000  "  (970  K) 
S  +  O3=  SO.,  -f  71000  "  (710  K) 

*  A  very  full  table  of  heats  of  formation  expressed  in  terms  of  K  will  be  found 
in  Ostwald's  Outlines  of  General  Chemistry. 


CALCULATIONS   OF    CALORIFIC    POWER.  Ill 

We  see  from  these  the  amounts  of  heat  given  up  by  the 
formation  of  18  grams  of  water,  28  grams  of  carbonic 
oxide  etc.  For  the  purpose  of  calculating  the  calorific 
power  of  fuel,  it  will  be  more  convenient  to  have  the 
amount  of  heat  formed  by  the  combustion  of  one  gram 
of  hydrogen,  carbon  and  sulphur.  If  2  -|-  grams  of  hydro- 
gen and  1 6  grams  of  oxygen  give  18  +  grams  of  water 
and  68,400  calories :  one  gram  of  hydrogen  will  give 

— ^ —  or  34,200  calories  :   similarly  one  gram  of  carbon 

—  or  8083  calories  and  one  gram  of  sulphur  221 

=  2219  calories. 

These  heats  of  formation,  so  essential  to  the  calcula- 
tion of  calorific  power,  are  given  various  figures  by  dif- 
ferent observers   and  it   is   a  difficult  matter  to  decide 
which  to  use.     We  shall  adopt   the  following  as  having 
the  greatest  probability  of  accuracy  and  at  the  same  time 
being  in  most  general  practical  use. 
One  gram  carbon  to  carbonic  acid  8080  calories 
One  gram  hydrogen  to  water  (condensed)  34500  calories 
One  gram  sulphur  to  sulphur  dioxide  2  2 20  calories 
These  numbers  are  the  same  in  greater  calories  when 
one  kilogram  of  the  element  is  burned. 

It  will  be  seen  from  the  preceding  variations  between 
the  values  of  the  fundamental  data,  that  considerable  un- 
certainty exists  concerning  thermo-chemical  figures. 
The  relative  accuracy  is  much  greater  than  the  absolute; 
so  the  data  employed  should  always  be  taken  from  the 
same  source  and  if  possible  obtained  by  the  same 
method. 


112  CALORIFIC    POWER   OF    FUEL. 

With  this  introduction  we  proceed  to  the  consider- 
ation of  the  calculations  of  the  calorific  power  of  fuel 
from  a  practical  standpoint. 

CALORIFIC    POWER    OF    FUEL. 

The  calorific  power  of  any  solid  fuel  such  as  coal  can 
be  determined  by  three  methods. 

i  st.   By  combustion  in  a  calorimeter. 

2nd.  By  calculation,  from  the  results  of  an  ultimate 
analysis. 

3rd.  By  calculation,  from  the  weight  of  lead  reduced 
from  litharge  by  fusion  with  a  weighed  quantity  of  the 
coal. 

Only  the  first  and  second  methods  are  applicable  to 
liquids  or  gases. 

i  st.  Of  these  three  methods  the  first  is  the  most  ac- 
curate, as  it  affords  a  direct  measure  of  the  heating 
power  of  the  fuel  when  burned  under  the  most  favorable 
conditions.  The  calculations  involved  do  not  specially 
belong  to  analytical  chemistry,  so  the  reader  is  referred 
to  Poole  on  the  Calorific  Power  of  Fuels,  for  both  a  de- 
scription of  the  apparatus  and  directions  for  the  calcula- 
tions. 

2nd.  The  second  method  is  largely  used  and  is  im- 
portant for  analysis,  as  they  usually  have  the  apparatus 
required  to  get  the  analytical  results,  while  they  are  not 
often  supplied  with  an  expensive  calorimeter. 

For  commercial  purposes,  calorific  power  is  expressed 
in  four  ways  : 

ist.  In  calories  per  kilogram  (greater  calories),  for 
example  7000  calories  means  that  one  kilogram  of  the 
fuel  will  raise  the  temperature  of  7000  kilograms  of 
water  one  degree  centigrade. 


CALORIFIC    POWER    OF    FUEL.  113 

2nd.  In  British  thermal  units  (B.T.U.)  or  the  amount 
of  heat  necessary  to  raise  one  pound  of  water  one  degree 
Fahrenheit. 

3rd.  In  kilograms  of  water  evaporated  per  kilogram 
of  coal. 

4th.   In  pounds  of  water  evaporated  per  pound  of  coal. 

The  relations  between  these  various  ways  of  express- 
ing calorific  power  are  not  complicated.  To  convert 
calories  per  kilogram  into  British  thermal  units  it  is  only 
necessary  to  multiply  by  1.8  for  if  a  kilogram  of  coal 
heats  7000  kilograms  of  water  one  degree  centigrade, 
one  pound  of  coal  will  heat  7000  pounds  of  water  one  de- 
gree centigrade,  and  it  only  remains  to  multiply  by  the 
factor  for  converting  centigrade  to  Fahrenheit  degrees. 
Conversely  to  change  B.T.U.  to  calories  divide  by  1.8 
or  multiply  by  5/9. 

To  find  the  number  of  kilograms  of  water  evaporated 
by  a  kilogram  of  fuel,  divide  the  calorific  power  (in 
calories  per  kilogram)  by  536,  the  number  of  calories  re- 
quired to  convert  one  kilogram  of  water  at  100°  C  into 
steam  at  100°  C.  If  it  should  be  necessary  to  raise  the 
water  from  some  temperature,  for  example  15°  C  to 
100°  C  and  then  evaporate  it :  divide  by  536  +  ( 100  -  1 5) 
or  621.  As  the  specific  heat  of  water  is  taken  as  unity. 

To  find  the  number  of  pounds  of  water  evaporated  by 
a  pound  of  fuel,  divide  the  calorific  power  expressed  in 
B.T.U.  by  965  (536  X  1.8).  For  example,  how  many 
pounds  of  water  will  be  heated  from  59°  F.  to  212°  F. 
and  evaporated  by  one  pound  of  a  coal  whose  calorific 
power  is  7000  (calories  per  kilogram)  ?  7000  X  1.8  = 
12,600  B.T.U.  It  requires  212-59  =  153  B.T.U. 
to  heat  a  pound  of  water  from  59°  F.  to  212°  F.  and 
965  B.T.U.  to  convert  it  into  steam  at  212°  F:  So, 
i  2600  -=-  (153  -f-  965)  =  1 1. 27  pounds. 


114  CALORIFIC    POWER    OF   SOLIDS. 

CALORIFIC    POWER    OF    SOLIDS. 

Given  the  composition  of  perfectly  dry  oak  wood : 
carbon,  50.16%:  hydrogen,  6.02%:  oxygen,  43.36%: 
nitrogen,  0.09%  :  ash,  0.37%  :  to  calculate  its  calorific 
power. 

One  kilogram  will  contain  : 

Carbon       501.6  grams  Nitrogen  0.9  grams 

Hydrogen    60.2       "  Ash  3.7 

Oxygen      433.6       " 

As  part  of  the  hydrogen  is  already  in  combination 
with  oxygen,  only  the  excess  over  the  amount  necessary 
to  form  water  with  all  the  oxygen  is  available  as  fuel : 
as  1 6  parts  of  oxygen  combine  with  2  of  hydrogen,  the 
hydrogen  already  in  combination  with  oxygen  will  be 
433.6  -r-  8  =  54.2  grams:  So  we  have  available  as  fuel 
6  grams  of  hydrogen  and  501.6  grams  of  carbon. 

0.006  kilogram  hydrogen    X  34500  =      207  calories 
0.5016        "          carbon        X     8080  =  4053        " 

Theoretical  heating  power  =  4260  calories. 
If  the  products  of  combustion  escape  at  a  temperature 
of  or  above  ioo°C.,  the  heat  given  up  by  the  condensa- 
tion of  the  steam,  which  is  lost,  must  be  subtracted. 
This  is  found  as  follows  :  60.2  grams  of  hydrogen  give 
9X  60.2  =  541.8  grams  of  water;  and  as  536  calories 
are  required  to  convert  one  kilogram  of  water  at  100°  to 
steam  at  100°,  the  heat  lost  in  this  case  will  be  equal  to 
that  required  to  change  541.8  grams  of  water  into  steam 
at  ioo°C  or  536  X  0.5418  =  290  calories  :  so  the  avail- 
able heat  becomes  3970  calories.  When  the  fuel  has 
also  hygroscopic  moisture  this  must  also  be  evaporated 
and  if  the  products  are  not  condensed,  the  heat  neces- 
sary to  do  this  must  be  deducted. 


CALORIFIC    POWER   OF   SOLIDS.  Il-> 

To  calculate  the  calorific  power  of  a  coal  whose  com- 
position is:  carbon,  80.55%;  hydrogen,  4.50%;  sul- 
phur, 0.54%;  oxygen,  (by  difference)  2.70%;  nitrogen, 
i.  08%;  moisture,  2.92%;  ash,  7.71%. 

The  nitrogen  and  ash  are  inert  and  simply  diminish 
the  available  calorific  power  by  the  heat  expended  in 
bringing  them  to  the  temperature  of  the  products  of  com- 
bustion, this  amount  is  relatively  very  small  and  is  neg- 
lected. 

The  moisture  is  objectionable  as  it  must  be  raised  to 
ioo°C  and  then  evaporated,  thus  rendering  latent  a  por- 
tion of  the  heat  evolved  which  can  however  be  recovered 
if  the  products  of  combustion  are  cooled  below  ioo°C. 

The  available  hydrogen  is  calculated  as  before, 


One  kilogram  of  the  coal  contains  available  as  fuel  : 
41.6  grams  of  hydrogen  giving  1435.2  calories 
805.5        "       "  carbon  "       6508.4 

5.4        '•       "sulphur         "  12.0          " 

Theoretical  calorific  power  =  7943.6          " 

This  on  the  basis  of  the  condensation  of  the  aqueous 
vapor  produced  which  will  then  give  up  heat  equal  to 
that  required  to  vaporize  it,  so  no  deduction  is  made.  If 
the  products  are  allowed  to  escape  without  condensation 
of  the  aqueous  vapor,  as  is  always  the  case  in  practice  ; 
the  number  of  calories  necessary  to  convert  into  vapor 
all  the  water  present  must  be  subtracted. 

This  will  be,  9  X4-5  (the  percentage  of  hydrogen)  = 
405  grams  per  kilogram  plus  moisture  29.2  grams  = 
434.2  grams  or  0.4342  kilogram;  and  0.4342  X  53^ 
(the  number  of  calories  necessary  to  convert  one  kilo- 


116  CALORIFIC    POWER   OF   SOLIDS. 

gram  of  water  at  ioo°C  to  steam  at  ioo°C)  gives  232.7 
calories.  So  under  these  conditions  the  available  calori- 
fic power  of  the  coal  is  7710.9  calories.  This  can  be 
corrected  further  for  the  heat  necessary  to  raise  the  fuel 
to  ioo°C  and  for  the  heat  lost  by  the  escape  of  the  pro- 
ducts of  combustion  at  temperatures  above  ioo°C  &c. 

Taking  7711  as  the  calorific  power  of  the  coal,  let 
us  see  the  quantities  of  water  it  will  evaporate. 

Kilograms  of  water  at   ioo°C  per  kilogram  of  coal 

will  be   7711    =  14.38. 

536 
Kilograms  of  water  at   i5°C   per  kilogram  of  coal 


will  be 7711 


-.-  =  12.41. 


536+  (100-  15) 
Pounds  of  water  at  2i2°F.  per  pound  of  coal 

will  be  7_7IU<_±?  or   77H  =  I4.38. 

965  536 

Pounds  of  water  at  4O°F.  per  pound  of  coal 
will  be        77"   xj.8       =  ^^ 

965  +  (212-40; 

This  discussion  can  be  summed  up  by  the  following 
formulae.  Let  x  be  the  calorific  power  of  the  fuel,  and 
let  C,H,O,S  and  Wbe  the  weights  of  carbon,  hydrogen, 
oxygen  and  hygroscopic  water,  expressed  in  kilograms: 
then  we  have, 

x  =  8o8oC  +  34500  (H  -  -?-)  +  2220  S. 

8 

An  expression  for  the  theoretical  calorific  power.  And, 
x  =  8o8o  C  +  34500  (H-(5)+222oS-536(W  +  9H) 

o 

For  the  available  calorific  power  when  the  products 
of  combustion  are  not  condensed.  The  heat  generated 


CALORIFIC    POWER    OF    LIQUIDS.  117 

by  the  sulphur  can  usually  be  omitted  without  materi- 
ally affecting  the  results. 

Numerous  formulae  are  given  by  different  authori- 
ties, which  differ  somewhat  from  the  preceding  and 
attempt  to  bring  the  results  by  calculation  nearer  those 
obtained  by  the  calorimeter,  or  the  conditions  of  actual 
practice.  '  Those  given  agree  very  closely  with  that  of 
Dulong  as  recommended  by  the  committee  of  the  Amer- 
ican Chemical  Society.* 

The  reader  is  left  to  adapt  them  to  particular  con- 
ditions, with  the  caution  not  to  trust  too  far  to  absolute 
figures  obtained  by  calculation.  The  use  of  the  method 
is  to  compare  one  fuel  with  another,  in  order  to  deter- 
mine which  is  more  economical,  and  for  this  purpose  it  is 
of  value. 

CALORIFIC    POWER    OF    LIQUIDS. 

These  present  no  difficulties  not  already  dicussed ;  for 
example,  to  calculate  the  theoretical  calorific  power  of 
methyl  alcohol,  CH;5OH.  We  find  the  percentage  com- 
position to  be  :  carbon,  37.46;  hydrogen,  12.58  ;  oxygen, 
49.95  :  or  expressed  in  kilograms  C,  0.3746  ;  H,  0.1258; 

O,  0.4995  :  tne  available  hydrogen  (H  -  — )  is  0.0634  so 

8 

we  have 

Hydrogen  0.0634  X  34500  =  2187.3  calories 
Carbon       0.3746  X    8080  =  3026.7  calories 
Theoretical  calorific  power         5214.     calories 
If  the  water  formed  is  not  condensed  we  must  deduct 
536  X  9  X  0.1258  =  607  calories.     Giving  4607  as  the 
available  calorific  power. 


*  I.  Am.  Chem.  Soc.   Dec. 


118  CALORIFIC    POWER   OF   GASES. 

The  theoretical  calorific  power  obtained  by  Stohmann 
is  5265  calories;  this  is  a  rather  remarkable  agreement 
as  the  calculated  calorific  power  of  liquids  is  not  often 
reliable. 

CALORIFIC    POWER    OF    GASES. 

In  connection  with  the  calorific  power  of  gases  some 
reasons  can  be  shown  for  the  discrepancy  in  results  be- 
tween the  calculated  and  actual  calorific  power  of  solids 
and  liquids.  Let  us  consider  the  thermo-chemical  re- 
actions : 

C  +  O2  =  CO,     +  970  K 
2HS+    O,  =  2H30  +  2(684)K 

C   +  2H,  =  CH4     +  218  K  hence 
CH4+  2O,  =  CO,,    +2H2O  +  2120  K 

2I2O  =   9/0    -f-    2(684)  -  2l8, 

or  the  heat  produced  by  the  combustion  of  16  grams  of 
CH4is  .2120  K;  less  by  218  K  than  the  heat  produced 
by  the  burning  of  i  2  grams  of  carbon  and  four  grams  of 
hydrogen. 

In  solids  and  liquids  when  we  do  not  know  the  chemi- 
cal compounds  and  only  the  ultimate  composition  we  may 
be  led  into  similar  errors. 

Sixteen  grams  of  methane  (CH4)  gave  2120  K  hence 
one  gram  gives  132.5  K  or  13250  calories  and  one  kilo- 
gram the  same  number  of  greater  calories.  In  the  table 
at  the  back  of  this  book,  we  find  the  weight  of  a  liter  of 
methane  to  be  0.71502  gram.  The  weight  of  a  cubic 
meter  is  1000  times  this  or  an  equal  number  of  kilograms. 
The  calorific  power  of  a  cubic  meter  will  therefore  be 
13250  X  0.715  =  9474  calories  (greater),  and  as  a  cubic 
meter  =  35.3171  cubic  feet,  the  calorific  power  per  cubic 
foot  is  268.2  calories.  To  convert  this  to  B.T.U.  per 


CALORIFIC    1'OWER  'OF    GASES.  119 

•cubic  foot  it  must  be   multipled  by  3.97  giving    1065  as 
the  calorific  power  of  methane  per  cubic  foot. 

Now  that  we  see  the  relation  of  these  values  we  can 
conveniently  use  the  following  figures  to  calculate  the 
theoretical  calorific  power  of  a  gas  per  cubic  foot  meas- 
ured at  o°  C  and  760  m.m.  pressure 

Hydrogen,  H2        347-2    B.T.U  per  cubic  foot. 

Carbonic  oxide  CO.      341.4        "          "        " 

Methane  CH4     1065 

Illuminants  C2H4  -f-  C;{H6  2000  B.T.U.  per  cubic  foot. 
If  we  assume  the  initial  temperature  of  the  gas  to  be 
60°  F.  and  the  products  to  escape  at  328°  F.,  without 
condensation,  these  figures  become  :  * 

Hydrogen  H2        264.      B.T.U.  per  cubic  foot 

Carbonic  oxide  CO       307.  "       " 

Methane  CH4     853. 

Illuminants  C3H4 -f  C3H6 1 700     " 

Given  a  gas  of  the  following  composition,  to  calculate 
its  theoretical  calorific  power  in  B.T.U.  per  cubic  foot: 
hydrogen,  50.4%:  methane,  33.7%:  illuminants, 
5.2%:  carbonic  acid,  2.6%:  carbonic  oxide,  5.8%: 
oxygen,  0.8%:  nitrogen,  1.5%. 

As  all  gases  expand  equally  the  percentage  compo- 
sition by  volume  is  not  affected  by  the  temperature  at 
which  the  analysis  was  made.  In  this  case  the  oxygen 
cannot  be  in  combination  with  the  hydrogen,  so  we  have 
available  as  fuel : 

Hydrogen    0.504  Cu.  ft.  X  347-2  =  174.98  B.T.U. 
Methane      0.337         "     X  1065.  =  358.90 
Illuminants  0.052         "     X  2000    =  104.00       " 


For  details  of   these  calculations  see    Stillman's    Engineering  Chemistry, 


120      CALORIFIC  POWER  FROM  WEIGHT  OF  LEAD  REDUCED. 

Carbonic  oxide  0.058  Cu.  ft.  X  341.4  =  19.80  B.T.LL 

Theoretical  calorific  power  per  Cu.  ft  657.68  B.T.U. 
To  convert  this  to  calories  per  cubic  meter  multiply  by 
8.896:  which  gives  5851. 

The  available  calorific  power,  under  the  conditions 
stated,  is  found  by  using  the  values  in  the  second  table. 

Many  corrections  might  be  added  and  other  calcu- 
lations, such  as  the  volume  of  air  required,  the  volume  of 
the  products  of  combustion,  calorific  intensity  &c.  Some 
of  these  can  be  done  by  the  methods  given  under  the 
calculations  of  gases,  the  others  seem  out  of  the  sphere 
of  analytical  chemistry. 

CALORIFIC    POWER    FROM    WEIGHT    OF    LEAD     REDUCED. 

The  method  of  calculating  calorific  power  from  the 
weight  of  lead  reduced  from  litharge  by  a  gram  of  the 
fuel  is  only  suitable  for  coal  or  coke  and  is  only  valuable 
when  the  average  weight  of  three  lead  buttons  is  multi- 
plied by  a  factor  proper  for  the  fuel. 

This  factor  should  be  obtained  by  comparison  with 
the  calorimeter  and  will  be  the  calories  developed  by 
one  gram  divided  by  the  lead  reduced  from  litharge  by 
one  gram.  When  this  is  once  obtained  it  will  give  sat- 
isfactory results  with  similar  coals. 

Berthier's  idea  that  the  calorific  power  was  directly 
proportional  to  the  oxygen  consumed,  i.  e.  the  lead  re- 
duced, is  disproved  by  the  following  : 

C+  2  PbO  =  2  Pb+  CO2 

12  parts  of  carbon  reduce  414  parts  of  lead  hence, 
i  kilogram  of  carbon  reduces  34.5  kilograms  of  lead, 
i  kilogram  of  carbon  gives  8080  calories,  hence, 

the  lead  reduced  times—     D   or  234.2  gives  the  calorific 

34-5 


CALORIFIC  POWER  FROM  WEIGHT  OF  LEAD  REDUCED.      121 

power.     The  factor  for  hydrogen  is  obtained  similarly  : 

H2+  PbO  =  Pb  +  H2O. 

i  kilogram  of  hydrogen  would  reduce  103.5  kilograms 
of  lead,  and  as  one  kilogram  gives   34500  calories  the 

factor  is   345°°  or  333.33. 

103-5 

So  it  is  evident  that  the  more  volatile  matter,  hydro- 
carbons &c.,  the  coal  contains,  the  higher  must  be  the 
factor  used. 

When  no  better  means  of  determining  calorific  power 
is  available  the  weight  of  lead  reduced  by  one  gram  times 
240  for  an  anthracite  or  times  268.3*  fora  bituminous 
coal,  will  give  approximately  the  calorific  power. 


*  Noyes  J.,  Am.  Chem.  boc.  Vol.  17,  848. 


EXAMPLES. 


1 .  Find  the  theoretical  calorific  power  of  wood,  whose 
composition  is:  carbon,  40%;  hydrogen,  4.8%;  oxygen, 
33.8%;  nitrogen,  0.4% ;  ash,  i%;    hygroscopic  water, 
20%.     Assuming  the  wood  to  be  at   ioo°C  and  that 
the  aqueous  vapor  is  condensed. 

Ans.     3430.4  calories. 

2.  Find  the  calorific  power  of  peat,  whose  composi- 
tion is:  carbon,  39.3%;   hydrogen,  3.9%;  oxygen,  23.2 
%;  ash.  7.3%;   water,  26.3%.     Assuming  that  the  pro- 
ducts  of  combustion   escape  at  ioo°C.  without  conden- 
sation. Ans.     3191.3  calories. 

3.  Find  the    theoretical  calorific  power   of  a   coal, 
whose  composition  is:   carbon,  83.75%;   hydrogen,  4. 13 
%•  oxygen.  2.65%;  nitrogen,  0.85%;   sulphur,  0.57%; 
water,  0.80%;  ash,  7.25%.       Ans.     8090.6  calories. 

4.  In   the  preceding    example,   how    many    calories 
must  be  deducted  if  the  aqueous  vapor  is  not  condensed? 

Ans.  .  203.5. 

5.  Find  the  theoretical  calorific  power  in  B.T.U.  of 
a  coal  whose  composition  is :  carbon,  70.50%;  hydro- 
gen, 4.76%;    oxygen,  15.71%;    nitrogen,  1.36%;    sul- 
phur, 1.39%;  ash,  6.28%.  Ans.      12048. 

6.  How  many  kilograms  of  water  at  20° C   will  be 
evaporated  by  one  kilogram   of  a  coal,  whose  calorific 
power  is  13535  B.T.U.  ?  Ans.      12.2 

(122) 


EXAMPLES,  123 

7.  How  many  pounds  of  water  at  2$°C  will  be  eva- 
porated by  a  coal  whose  calorific  power  is  8000  ? 

Ans.      13.1 

8.  How  many   kilograms  of  water  at  60° C  will  be 
evaporated   by  a  coal  whose   calorific  power  is    11270 
B.T.U?  Ans.   10.86 

9.  What  is    the    difference  between   the  calculated 
available  calorific  powers  of  ethyl  and  methyl  alcohol. 

Ans.      2000  calories  (in  round  numbers). 

10.  What   is    the  calorific  power  of  a  natural  gas, 
whose  composition  is :    methane,    96.50%;    illuminants, 
i%;   carbonic  oxide,  0.5%;    oxygen,  2%  ;      in  B.T.U. 
per  cubic  foot?     Gas  and  products  at  o°C. 

A  ns.      1 049. 

11.  What  is  the  calorific  power  of  a  coal  gas,  whose 
composition  is:     hydrogen,    50.4%;     methane,  33-7%; 
illuminants,  5-2%;  carbonic  acid,  2.6%;  carbonic  oxide, 
5.8%;    oxygen,  0.8%;  nitrogen,  1.5%;    in  B.T.U.  per 
cubic  foot  ?      Assume  that  the  gas  enters  at  6o°F  and  the 
products  of  combustion  escape  at328°F. 

Ans.     527. 

12.  What  is  the  calorific  power  of  the  following 
water  gas:  Hydrogen,  55%;  methane,  5.75%;  illumin- 
ants, 1.20%;  carbonic  acid,  4.05%;  carbonic  oxide, 
31.00%;  nitrogen.  3%;  in  calories  per  cubic  meter? 
Gas  and  products  at  o°C.  Ans  3398. 


CHAPTER  X. 

ELECTRIC     AND    ELECTROLYTIC    CALCULATIONS    FOR 
DIRECT    CURRENTS. 

The  calculations  described  in  this  chapter,  although 
not  essential  to  analytical  chemistry,  are  added  because 
electricity  is  so  largely  employed  for  both  laboratory 
and  industrial  operations,  and  it  is  advisable  that  a 
chemist  be  able  to  make  the  ordinary  calculations  invol- 
ved in  its  use.  In  order  to  make  the  description  as  con- 
cise as  possible  the  subject  is  treated  differently  from  the 
preceding  chapters,  and  the  important  laws  are  expres- 
sed by  formulae,  many  of  which  are  taken  from  an  article 
by  Prof.  Crocker  on  the  Theory  of  Electro -metallurgy.* 

ELECTRIC    CALCULATIONS. 

The  ampere  is  the  unit  of  current,  that  is  the  rate  of 
flow ;  it  is  the  strength  of  current  which  when  passed 
through  a  solution  of  silver  nitraie  in  water,  under  cer- 
tain standard  conditions,  will  deposit  silver  at  the  rate  of 
0.001118  gram  per  second.  Five  amperes  will  deposit 
five  times  this  amount  every  second.  So  to  deposit 
1.3416  grams  of  silver  from  an  aqueous  solution  of  silver 
nitrate  in  ten  minutes  requires  two  amperes ;  for 

1-3416  2> 

0.001118  x  600 

The  coulomb  or  ampere-second  is  the  unit  of  quan- 
tity. It  is  the  quantity  of  electricity  which  flows  along  a 


School  of  Mines  Quarterly.     January,  1895. 
(134) 


ELECTRIC   CALCULATIONS.  125 

conductor  in  one  second  when  the  current  is  one  ampere. 
The  total  quantity  or  number  of  coulombs  flowing  in  a 
given  time  is  equal  to  the  strength  of  current  in  amperes 
multiplied  by  the  time  in  seconds  or  expressed  in 
symbols 

Q  =  ct  (,) 

when  Q   =  quantity  in  coulombs 

C   =  strength  of  current  in  amperes 
t  =  time  in  seconds 

In  dealing  with  large  quantities  of  electricity  the 
ampere-hour  is  used  as  a  unit,  an  ampere  flowing  for  an 
hour.  This  is  3600  coulombs. 

If,  for  example,  1.5  amperes  flow  for  45  minutes  the 
number  of  coulombs  that  have  passed  will  be 

15  X  45  X  60  =  4050  equal  to  1.125  ampere-hours. 

The  ohm  is  the  unit  of  resistance.  It  is  the  resist- 
ance offered  to  an  unvarying  electric  current  by  a  col- 
umn of  mercury  at  o°C,  one  square  millimeter  in  cross 
section  and  106.3  centimeters  long. 

The  resistance  of  a  conductor,  whether  solid  or 
liquid,  to  the  passage  of  electricity  has  been  found  to  de- 
pend upon  its  dimensions  and  the  material  of  which  it 
consists.  For  a  given  material  it  varies  directly  as  the 
length  and  inversely  as  the  area  of  the  cross  sec- 
tion. 

The  specific  resistance  of  any  material  is  the  resist- 
ance measured  between  opposite  faces  of  a  cube  of  the 
material  one  square  centimeter  in  area. 

The  resistance  of  any  conductor  is  given  by  the 
formula, 

R  =    p  1  (2) 


126  ELECTRIC   CALCULATIONS. 

in  which 

R  is  the  resistance  of  the  conductor  in  ohms 

p  is  the  specific  resistance  of  the  material  in  ohms 

1    is  the  length  in  centimeters,  and 

a    is  the  area  of  the  cross-section  in  square  centi- 
meters. 

Therefore,  if  the  specific  resistance  of  a  conductor  is 
known  the  total  resistance  can  be  calculated  from  its  di- 
mensions. To  find  the  resistance  of  200  meters  of  Ger- 
man-silver wire  having  a  cross  section  of  10  square 
millimeters;  given  the  specific  resistance  of  German - 
silver  as  0.0000209.*  Reduce  the  area  to  square  centi- 
meters and  the  length  to  centimeters  and  the  resistance 

2OOOO 

is  0.0000209  X   -    =4.18  ohms. 

10 

Similarly,  if  a  glass  tube,  20  centimeters  long  with 
an  internal  cross  section  of  2  square  centimeters,  is  filled 
with  a  solution  of  zinc  sulphate  whose  specific  resistance 
is  30  ohms,  the  total  resistance  will  be 

^s    2O  , 

30  X  —  =  300  ohms. 

The  specific  resistance  is  not  absolutely  constant  but 
changes  with  the  temperature.  In  the  case  of  all  metals 
it  increases  with  the  temperature  while  for  carbon,  selen- 
ium, liquids  and  solutions  it  diminishes  as  the  tempera- 
ture rises.  For  aqueous  solutions  it  naturally  varies 
with  their  strength. 

Conductors  are  connected  in  series  when  the  current 
passes  through  one  after  the  other;  in  parallel,  when 
they  are  connected  side  by  side  so  as  to  furnish  more 
than  one  path  for  the  current.  The  total  resistance  of  a 
number  of  separate  resistances  connected  in  series  is 

*  See  Tables. 


ELECTRIC  CALCULATIONS.  127 

equal  to  their  sum.  If  two  paths  of  equal  resistance  are 
offered,  the  total  resistance  is  only  one  half,  as  this  is 
equivalent  to  doubling  the  cross  section  of  the  conduc- 
tor, or  the  resistance  of  a  number  of  equal  resistances  in 
parallel  is  equal  to  the  resistance  of  one  divided  by  the 
number  so  connected. 

Conductance  is  the  reciprocal  of  resistance,  or  the 
ease  with  which  a  current  is  allowed  to  pass.  The  total 
conductance  of  a  number  of  separate  conductances  con- 
nected in  parallel  is  equal  to  their  sum.  Thus  five  con- 
ductors each  of  50  ohms  resistance,  connected  in  series 
give  a  resistance  equal  to  their  sum,  2 50  ohms;  while 
connected  in  parallel  they  give  the  resistance  of  one 
divided  by  the  number,  10  ohms.  If  three  resistances  of 
20,  50  and  80  ohms  respectively,  are  connected  in  series 
their  total  resistance  is  1 50  ohms ;  while  if  connected  in 
parallel  their  resistance  is  12.12  ohms;  for  their  respective 

conductances  will  be  — »  —  and  —  and  their  total  con- 
20   50          80 

ductance  the  sum  of  the  three,  0.0825.  So  that  their 
total  resistance  is  the  reciprocal  of  0.0825  or  12.12  ohms. 

The  volt  is  the  unit  of  electromotive  force  or  elec- 
trical pressure. 

The  relations  between  electromotive  force,  current, 
and  resistance  are  expressed  by  Ohms  Law;  that  the 
electromotive  force  acting  between  the  extremities  of  any 
part  of  a  circuit  is  the  product  of  the  strength  of  the  cur- 
rent and  the  resistance  of  that  part  of  the  circuit,  or 
E  =  C  R 

or  transposed  C  =  — —  (3) 

K. 

To  comply  with  Ohms  Law  and  the  definitions  of  the 
ampere  and  ohm  the  unit  of  electromotive  force,  the  volt 


128  ELECTRIC   CALCULATIONS. 

must  be  that  electromotive  force  which  steadily  applied 
to  a  conductor  whose  resistance  is  one  ohm  will  produce 
a  current  of  one  ampere. 

The  Clark  cell  at  15°  C  is  the  most  constant  source 
of  electromotive  force  and  is  used  as  a  standard.  It 
gives  an  electromotive  force  of  1.434  volts. 

Ohms  law,  equation  (3)  states  that  the  amperes  flow- 
ing in  any  circuit  are  equal  to  the  volts  divided  by  the 
ohms.  This  is  the  most  important  electrical  law  for  by 
it  when  any  two  of  the  quantities  involved  are  known 
the  third  may  be  determined.  For  instance  if  a  conduc- 
tor of  10  ohms  resistance  has  flowing  through  it  3  am- 
peres, the  voltage  measured  between  its  ends  must  be 
10X3  =3°  volts.  Again  if  18  volts  are  applied  to  the 
ends  of  a  conductor  of  9  ohms  resistance  the  current  will 
be  18  -r-  9  =  2  amperes.  Finally  if  o.  i  of  an  ampere  is 
desired  from  a  source  whose  electromotive  force  is  two 
volts,  the  total  resistance  of  the  circuit  must  be  20  ohms 

for  as  E  =  C  R.      R  =  —  and  2  -=-  o.  i  =20  ohms. 

We  are  now  in  a  position  to  consider  the  relations  be- 
tween electrical  energy,  heat  and  power. 

Joule  has  shown  by  experiment  that  the  relation  be- 
tween the  heat  developed  in  a  circuit  and  the  current 
flowing  through  it  is  expressed  by  the  equation 

V  =  C2RtXo.24  (4) 

When 

C  is  the  current  in  amperes 

R  is  the  resistance  in  ohms 

t    is  the  time  in  seconds  that  the  current  has  passed 

V  is  the  heat  in  calories. 


ELECTRIC   CALCULATIONS.  129 

Consequently  2  amperes  passing  through  5  ohms  for 
10  seconds  develope 

4  X  5  X  10  X  0.24  =  48  calories. 

The  constant  0.24  can  be  eliminated  from  equation 
(4)  if  instead  of  expressing  the  heat  developed  in  calo- 
ries we  measure  it  in  joules  :  for  the  joule,  the  electrical 
unit  of  heat,  is  0.24  of  a  calorie.* 

So  if  we  designate  the  heat  produced  in  joules  by  W 
the  expression  becomes 

W  =  C'Rt  (5) 

and  the  heat  developed  by  2  amperes  passing  through  5 
ohms  for  10  seconds  is  4  X  5  X  10  =  200  joules. 

Equation  (5)  may  be  written  W  =  C  t  +  C  R.  But 
we  have  already  seen  from  equation  (i)  that  Q  =  C  t 
and  from  Ohms  law  that  E  =  C  R ;  substituting  these 
values  we  get : 

W  =  QE  (6) 

This  equation  (6)  shows  that  the  heat,  measured  in 
joules,  generated  in  a  conductor  is  equal  to  the  number 
of  coulombs  that  have  passed  multiplied  by  the  voltage 
between  the  ends  of  that  conductor. 

For  example,  if  1000  coulombs  (10  amperes  for  one 
minute  and  forty  seconds)  have  passed  through  a  con- 
ductor, the  voltage  between  whose  ends  is  10 ;  there  have 
been  generated  in  that  time  1000  joules,  equal  to  238.1! 
calories. 

Since  heat  is  a  form  of  energy,  the  energy  applied 
to  the  conductor  expressed  in  joules  is  Q  E.  So  that  the 

*  The  latest  determinations  give  one  calorie  equal  to  4.2  joules  which  makes 
this  constant  0.2381. 

f  From  this  point  one  calorie  will  be  taken  as  4.2  joules. 


130  ELECTRIC   CALCULATIONS. 

rate  of  work  or  power  expressed  in  joules  per  second  is 
^ —  or  since  —  =  C  from  equation  (  i)  ;  the  power  ex- 
pressed in  joules  per  second  equals  C  E. 

A  joule  per  second  is  called  a  watt  so  if  we  desig- 
nate the  power  expressed  in  watts  by  P  we  have 

P  =  C  E  (7) 

or  the  number  of  watts  in  any  electrical  circuit  is  equal 
to  the  product  of  the  amperes  and  the  volts.  A  circuit 
in  which  4.2  amperes  are  flowing  at  a  pressure  of  100 
volts  is  performing  work  at  the  rate  of  420  watts,  cor- 
responding to  100  calories,  per  second. 

The  relations  between  the  electrical  and  mechanical 
units  of  power  are  as  follows : — 

Rowland  has  shown  that  it  requires  780  foot  pounds 
of  energy  to  heat  one  pound  of  water  one  degree 
Fahrenheit,  or  3.1  foot-pounds  to  heat  one  gram  of 
water  one  degree  centigrade,  one  calorie.  Since  one 
calorie  is  equal  to  4.2  joules,  one  foot-pound  is  equal  to 
1.356  joules  or  one  joule  equal  to  0.737  foot-pound. 

550  foot-pounds  therefore  equal  746  joules,  and  since 
550  foot-pounds  per  second  are  one  horse-power  and  a 
joule  per  second  is  a  watt,  746  watts  equal  one  horse- 
power. 

The  practical  unit  for  measuring  large  quantities  of 
electrical  power  is  the  kilowatt,  one  thousand  watts. 
This  is  equal  therefore  to  737  foot-pounds  per  second  or 
approximately  i%  horse-power. 

ELECTROLYTIC  CALCULATIONS. 

When  an  electric  current  passes  between  platinum 
plates  immersed  in  water  acidified  with  sulphuric  acid, 


ELECTROLYTIC   CALCULATIONS.  131 

water  is  decomposed,  oxygen  is  liberated  at  the  platinum 
anode  and  hydrogen  at  the  platinum  kathode :  anode 
and  kathode  are  the  names  given  to  the  electrodes  by 
which  the  current  enters  (anode)  and  leaves  (kathode) 
the  liquid.  There  is  no  liberation  of  hydrogen  or  oxy- 
gen except  at  the  contact  of  the  liquid  with  the  elec- 
trodes. This  chemical  decomposition  effected  by  the 
electric  current  is  called  electrolysis  and  the  bath  or  ma- 
terial which  undergoes  decomposition  the  electrolyte. 

Electrolytes  may  be  solids,  liquids  or  as  recent  ex- 
periments have  shown,  gases.  Silver  iodide  is  an  exam- 
ple of  a  solid  electrolyte,  while  as  liquid  electrolytes  we 
have  solutions  of  mineral  salts  and  acids  as  well  as  many 
fused  salts.  An  electrolyte  must  necessarily  be  a  con- 
ductor of  electricity  and  in  the  case  of  liquids  be  ionized. 
Liquids  such  as  pure  water,  ether,  and  carbon  bisulphide 
which  are  not  perceptibly  ionized  are  not  considered 
electrolytes. 

The  constituents  into  which  the  electrolyte  is  split 
up  by  the  current  are  called  ions ;  that  which  is  liberated 
at  the  anode  is  the  anion,  that  which  is  liberated  at  the 
kathode  the  kathion.  With  very  few  exceptions  an  ele- 
ment or  radical  is  always  liberated  at  the  same  elect- 
rode. Ostwald*  gives  the  following  classification  of 
ions — 

KATHIONS. 

Monovalent :  H  (in  acids)  K.  Na.  Li.  Cs.  Rb.  Tl. 
Ag.  NH4.  NH3R  to  NR4  (R  being  an  organic  radical) 
Cu  (cuprous),  Hg  (mercurous). 

Divalent :  Ca.  Sr.  Ba.  Mg.  Fe  (ferrous),  Cu  (cupric), 
Pb.  Hg  (mercuric),  Co.  Ni.  Zn.  Cd. 

*  Scientific  Foundations  of  Analytical  Chemistry  p.  53. 


132  ELECTROLYTIC   CALCULATIONS. 

Trivalent :  AL  Bi.  Sb.  Fe  (ferric)  and  some  of  the 
rarer  earth  metals. 

Tetravalent  Sn  (?).  Zr. 

ANIONS. 

Monovalent  :  OH  (in  bases)  F.  Cl.  Br.  I.  NO3.  C1O3. 
C1O4.  BrO4.  MnO4  (in  permanganates)  and  the  anions 
of  all  other  monobasic  acids  the  (acid  molecule  minus 
one  hydrogen  which  goes  to  the  kathode). 

Divalent:  S.  Se.  Te  (?).  SO4.  SeO4.  MnO4  (in  man- 
ganates),  and  the  anions  of  dibasic  acids. 

Tri-to  hexavalent:  The  anions  of  the  tri-to  hexaval- 
ent  acids.  Elementary  anions  with  a  valency  of  more 
than  two  are  not  known. 

The  quantity  of  the  ions  deposited  by  the  passage  of 
a  current  through  an  electrolyte  was  shown  by  Faraday 
to  be  very  simply  related  to  the  quantity  of  electricity 
which  passes. 

Faraday's  first  law  of  electrolysis  states:  that  the 
quantity  of  an  electrolyte  decomposed  by  the  passage  of 
a  current  of  electricity  is  directly  proportional  to  the 
quantity  of  electricity  which  passes  through  it.  So  as 
long  as  the  quantity  of  electricity  remains  the  same 
it  is  immaterial,  so  far  as  quantity  deposited  is  con- 
cerned, whether  the  e'ectricity  passes  as  a  very  intense 
current  for  a  short  time  or  as  a  very  weak  current  for  a 
long  time. 

Faraday's  second  law  of  electrolysis  states  :  that,  if 
the  same  quantity  of  electricity  passes  through  different 
electrolytes,  the  weights  of  the  different  ions  liberated  will 
be  proportional  to  the  chemical  equivalents  of  the  ions. 
Thus  if  the  same  current  passes  through  a  series  of 
electrolytes  from  which  it  liberates  as  ions  oxygen, 


ELECTROLYTIC   CALCULATIONS.  133 

hydrogen,  silver,  and  chlorine,  then  for  every  8  grams  of 
oxygen  evolved  there  will  be  liberated  1.008  grams  of 
hydrogen  107.92  grams  of  silver  and  35.45  grams  of 
chlorine. 

The  electrochemical  equivalent  of  a  substance  is  the 
weight  in  grams  deposited  by  the  passage  of  one 
coulomb  of  electricity.  Using  this,  Faraday's  laws  can 
be  comprised  in  the  statement :  that  the  number  of 
grams  of  an  ion  deposited  during  the  passage  of  a  cur- 
rent through  an  electrolyte  is  equal  to  the  number  of 
coulombs  that  have  passed  multiplied  by  the  electro- 
chemical equivalent  of  the  ion.  Since  one  ampere  per 
second,  from  the  definition  of  unit  of  current,  deposits 
0.001118  gram  of  silver  and  the  chemical  equivalent  of 
silver  is  107.92,  the  weight  of  any  other  element  liber- 
ated by  one  ampere  per  second  will  be  found  by  the 
proportion. 

o.ooi  1 18  :  x  :  :  107.92  :  A. 
in  which  A  is  the  chemical  equivalent  of  the  element,  or 

A  (o.ooi  1 1 8) 

x  =  — i J 

107.92 

Therefore  since  — =    0.00001036,    the    total 

107.92 

weight  deposited  must  be  : 

w  =  0.00001036  A  C  t.  (8) 

In  which — 

w  is  the  weight  in  grams  deposited. 

A  is  the  chemical  equivalent,  that  is  the  atomic 
weight  divided  by  the  valency  of  the  deposited  ion 

C  is  the  current  in  amperes  and 

t    is  the  time  in  seconds  that  the  current  flows. 


134  ELECTROLYTIC   CALCULATIONS. 

Expressing  the  time  in  hours  and  denoting  it  by  N, 
equation  (8)  becomes: 

w  =  0.0373  A  C  N.  (9) 

C  N  being  ampere-hours. 

By  transposing  this  equation,  we  get  an  expression 
for  the  amount  of  current  required  to  deposit  a  given 
weight  in  a  given  time. 

f*  26.8       W  /  X 

-ATT  (10) 

and  since  from  (7)  the  power  in  watts  (P)  is  equal  to 
the  product  of  the  amperes  times  the  electromotive  force 
(C  E)  we  get 


An  expression  for  the  power  in  watts  necessary  to  de- 
posit w  grams  of  a  metal  whose  chemical  equivalent  is 
A  in  N  hours  when  an  electromotive  force  E  is  used. 

To  illustrate  the  application  of  these  formulae  : 

The  amount  of  oxygen  liberated  during  the  decom 
position  of  acidulated  water  by  a  current  of  2  amperes 
lasting  15  minutes  is  given  by  equation  (8) 

w  =  0.00001036  X  8  X  2  X  900  =  1.492  grams. 

The  amount  of  copper  deposited  in  3  hours  by  a  cur- 
rent of  25  amperes  from  a  solution  of  copper  sulphate  is 
found  by  equation  (9),  to  be 

w  =  0.0373  X  3  !-8  X  25  X  3  =  8-895  grams. 

To  determine  the  number  of  amperes  that  are  necers- 
sary  to  deposit  500  grams  of  nickel  in  5  hours,  we  em- 
ploy formula  (10)  and  find 
c  =  26.8  x  500 
29-35  x  5 

The  power  necessary  to  deposit  48.8  grams  of  zinc 


COUNTER-ELECTROMOTIVE   FORCE.  135 

in  four  hours  with  a  voltage  of  four  is  found  by  substitu- 
in  equation  (i  i),  to  be  40  watts. 

p  =  26.8  x  48.8  x  4  =  ts 

32.7  x  4 

The  fact  that  elements  which  form  two  series  of  salts 
have  different  electro-chemical  equivalents  according  to 
the  solution  from  which  they  are  deposited,  must  be  borne 
in  mind  when  these  calculations  are  made.  For  example 
copper  forms  two  chlorides  Cu2Cl2  and  CuCl2.  In  the 
former  its  chemical  equivalent  is  twice  its  value  in  the 
latter :  hence  the  actual  weight  of  copper  deposited  is 
twice  as  great  when  a  solution  of  cuprous  chloride  is 
electrolysed  as  with  a  solution  of  cupric  chloride ;  the 
current  and  time  being  the  same. 

COUNTER-ELECTROMOTIVE    FORCE, 

When  a  molecule  is  formed  by  the  combination  of  one 
or  more  elements,  a  definite  amount  of  heat  is  evolved  or 
absorbed.  In  most  stable  compounds  heat  is  evolved  : 
to  decompose  such  a  molecule  energy  must  be  supplied 
which  is  equivalent  to  this  amount  of  heat.  This  energy 
can  be  supplied  by  an  electric  current.  To  separate 
metallic  zinc  from  zinc  sulphate  requires  energy  to  be 
given  to  the  solution  equivalent  to  the  quantity  of  heat 
produced  by  the  formation  of  zinc  sulphate.  Conse- 
quently when  an  electric  current  passes  through  an 
electrolyte,  part  of  the  electrical  energy  furnishes  the 
energy  necessary  to  overcome  the  chemical  forces  and 
only  the  remainder  produces  heat  according  to  Joule's 
law.  The  electrolytic  cell  has  therefore  a  counter-elec- 
motive  force  resisting  the  passage  of  or  opposed  to  the 
current  and  the  electromotive  force  producing  that  cur- 
rent. 


136  COUNTER-ELECTROMOTIVE    FORCE. 

Calling  this  counter-electromotive  force  e,  the  energy 
supplied  must  be   a  certain  number  of  joules  such  that 
the  electrical  energy  =  Qe  =  Get ;  depending  upon  the 
number  of  coulombs  that  have  passed. 
(Compare  equation  6) 

Or  expressed  in  calories, 

Electrical  Energy  =  —  Cet 

4.2 

Now  the  energy  of  combination  of  various  compounds 
may  be  measured  by  the  heat  which  is  liberated  when 
the  combination  takes  place,  that  is 

Chemical  Energy  =  w  h, 

when  w  is  the  weight  of  the  element  in  grams  and  h  is 
the  number  of  calories  produced  when  one  gram  of  this 
metal  combines  with  the  other  constituents  of  the  com- 
pound. 

Now  from  Faraday's  law  of  electrolysis  as  given  by 
equation  (8)  we  have  that  w,  the  weight  of  metal  de- 
posited =  0.00001036  A  C  t,  substituting  this  value  for 
w,  we  have 

Chemical  Energy  =  0.00001036  ACth         (12) 

This  chemical  energy  must  equal  the  electrical  energy 
supplied  to  the  cell  available  for  electrolysis,  that  is  the 
electrical  energy  not  producing  heat. 
Therefore 

— L_  Cet  =  0.00001036  ACth 
4.2^ 

which  by  cancelling  Ct  and  reducing  gives 

e  =  0.0000435  Ah  (13) 

So  the  counter  electromotive  force  is  equal  to  the  heat 
of  combination  of  one  gram  in  calories  times  the  chemi- 
cal equivalent  times  a  constant,  0.0000435. 

Instead  of  stating  the  heat  of  combination  in  terms  of 


COUNTER-ELECTROMOTIVE    FORCE.  137 

one  gram  of  the  metal,  the  more  convenient  method  for 
electrolytic  calculations  is  to  state  it  in  terms  of  the 
chemical  equivalent  in  grams.  For  example,  the  chemi- 
cal equivalent  of  zinc  is  32.7  so  that  the  heat  of  combi- 
nation of  32.7  grams  of  zinc  with  other  substances  is  the 
value  to  be  used.  When  the  values  are  given  in  terms 
of  the  molecular  weight  in  grams  these  must  be  reduced 
to  the  heats  of  combination  per  equivalent  by  dividing 
by  the  valency  or  multiple  of  the  valency  of  the  element. 
The  heat  of  formation  of  A12(SO4)3  is  given  by  Thomp- 
son as  150630  calories:  this  means  the  combination  of 
54.22  grams  of  aluminium  with  sulphur  and  oxygen  to 
give  342.43,  the  molecular  weight  in  grams  of  aluminium 
sulphate.  As  the  equivalent  of  aluminium  is  9.037  we 
must  divide  by  6  (twice  the  valency  as  there  are  two 
atoms)  to  obtain  the  heat  per  equivalent,  25315  calories. 
Calling  the  heat  per  chemical  equivalent  in  grams 
H  and  remembering  that  A  is  the  chemical  equivalent 
and  h  is  the  heat  evolved  by  the  combination  of  one  gram, 

H  =  Ah, 

substituting  this  value  in  equation  (13)  we  get 
e  =  0.0000435  H  and 

since    0.0000435    equals -  this    equation    reduces 

22988 

without  appreciable  error  to  the  very  simplest  form  of 
Thompson's  law. 

e  =  JL_  (H) 

23000 

A  definite  case  may  make  this  deduction  clearer. 
When  31.8  grams  of  copper,  (the  weight  of  the  chemical 
equivalent  in  grams)  combines  to  form  copper  sulphate, 
27980  calories  are  developed.  Now  if  no  energy  is  lost 
as  heat  the  electrical  energy  necessary  to  deposit  31.8 


138  COUNTER-ELECTROMOTIVE   FORCE. 

grams  of  copper  from  copper  sulphate  must  equal  this. 
This  electrical  energy  is  the  product  of  e,  the  counter- 
electromotive  force,  and  Q,  the  number  of  coulombs  ne- 
cessary to  deposit  31.8  grams  of  copper.  We  find  the 
number  of  coulombs  by  substituting  in  equation  (8) 
giving 

31.8  =  0.00001036  X  31-8  X  Q 

or  Q  =  -  =  9652^  coulombs. 
0.00001036 

Now  as  one    calorie   equals  4.2  electrical  heat  units 
we  have 

27980  X  4-2  =e  X  96525 

or  e  .  27980^4.2  or         ticall     e  =  27980  =  122 
96525  23000 

IT 

volts  agreeing  with  formula  (14)  e  =  -       -  where  H  is 


the  heat  produced  by  the  combination  of  the  chemical 
equivalent  in  grams  expressed  in  calories. 

This  exceedingly  simple  formula*  allows  the  elec- 
tromotive force  of  any  chemical  reaction,  whether  it  be 
direct  as  in  a  voltaic  cell,  or  counter  as  in  an  electrolytic 
cell,  to  be  easily  calculated  from  the  heats  of  combi- 
nation. The  calculated  values  agree  in  many  cases  very 
closely  with  those  found  by  experiment.  The  discrep- 
ancies probably  are  due  to  the  fact  that  all  the  energy 
of  combination  in  the  case  of  a  voltaic  cell,  for  instance, 
may  not  be  converted  into  electrical  energy,  as  some 
may  be  converted  into  heat  which  appears  in  the  cell. 

To  calculate  the  voltage  of  a  Daniel's  cell  :  zinc  is  dis- 
solved at  one  pole  to  form  zinc  sulphate,  the  electromotive 

*It  is  also  applied  to  fused  electrolytes.  See  Langley,  J.  Am.  Chem.  Soc., 
vol.  16,  p.  52,  1894. 


COUNTER-ELECTROMOTIVE    FORCE.  139 

force   is    (14)   53°45  =  2.31   volts  :    at  the  other   pole 

'   23000 

copper  is  deposited  which  sets  up  a  counter-motive  force 
found  in  the  same  way  to  be  1.22  volts.  So  that  the 
available  electromotive  force  of  the  cell  (neglecting  the 
slight  electromotive  force  where  the  two  solutions  touch) 
is  equal  to  2.31  —  1.2  2  or  1.09  volts.  The  electromotive 
force  determined  experimentally  is  about  1.08  volts. 
The  table  at  the  back  of  the  book  gives  some  of  the  most 
important  values  of  H  to  be  substituted  in  equation  14. 
If  other  tables  are  used,  the  different  methods  of  express- 
ing calorific  power  must  be  considered  also  in  some 
cases  the  heat  of  solution  and  that  of  any  chemical  com- 
binations which  take  place  in  the  cell  and  so  give  energy 
which  aids  the  current.  For  example  when  copper  sul- 
phate is  decomposed  in  an  aqueous  solution  we  have  : 
Kathode  ^-oCu  |  SO4°— >H2SO4*— °  H2  |  O  »-* Anode. 
We  have  in  reality  two  decompositions  and  one  com- 
bination so  the  electrical  energy  to  be  supplied  is  that 
equivalent  to  the  heat  of  combination  of  CuSO4  -f-  its 
heat  of  solution  in  water  +  the  heat  of  combination  of 
H2O  —  (the  heat  of  combination  of  sulphuric  acid  -{-  the 
heat  developed  on  diluting  with  water). 

When  the  heats  of  combination  are  taken  from  Ost- 
wald's  tables  H  is  obtained  as  follows  for  copper  sulph- 
ate and  nickel  sulphate 

CuSO4  i826K     NiSO4aq.       2294K 

Heat  of  solution       is8K     H2O  684JC 

H2O  684K  2978K 

2668K     H2SO4aq.       2ioj)K 
HoS04aq.  2IQ9K  869K 

559K 
As    these  values  are  based    on    molecular    weights    in 


140  OHM'S   LAW   FOR    ELECTROLYSIS. 

grams  to  be  converted  to  equivalent  weights  they  must 
each  be  divided  by  two,  since  copper  and  nickel  are 
divalent  ions ;  and  since  K  =  100  calories  we  get, 

2-^9v>  =  I>22  voits  for  CuSO4 
230 

and  434i5  =  1.89  volts  for  NiSO4 

230 

The  counter-electromotive  force  of  any  electrolytic  cell  can 
be  easily  determined  experimentally  as  follows :  Cause  some 
deposition  of  the  metal  to  take  place  by  sending  a  current 
through  the  cell  with  a  voltmeter  connected  between  the  elec- 
trodes. After  a  few  moments,  break  the  main  circuit,  and 
watch  the  voltmeter  closely ;  the  moment  this  occurs  the  needle 
will  fall  back  to  a  certain  point  and  stop  there  for  an  instant, 
then  it  will  gradually  drop  down  to  zero.  The  point  where  the 
voltmeter  needle  stops  first  is  the  counter-electromotive  force  of 
the  cell. 

OHM'S    LAW     FOR    ELECTROLYSIS. 

In  a  circuit  where  electrolytic  work  is  being  done 
the  quantities  of  electricity  are  not  related  in  the  form 
of  Ohm's  law  already  given  (3)  as  the  counter-electro- 
motive force  must  be  considered.  The  total  electrical 
energy  supplied  to  a  circuit  is  given  by  equation  (6)  as 
equal  to  QE  or  CEt. 

Similarly  the  chemical  work  done  is  Qe  or  Get. 

The  heat  generated  is,  from  equation  (5),  C2Rt. 
Now  from  the  principle  of  conservation  of  energy  we 
know  that  the  electrical  energy  supplied  must  equal  the 
sum  of  the  heat  produced  and  the  chemical  work  done. 
Hence 

CEt  =   C2  Rt  +  Get 
Cancelling  C  and  t  we  obtain 

E   =  CR  +  e.  (15) 


OHM'S    LAW   FOR    ELECTROLYSIS.  141 

AVhere  E  is  the  electromotive  force  required  to  produce 
the  current  C,e  the  counter-electromotive  force  of  the 
electrolytic  cell  and  R  the  total  ohmic  resistance  in  the 
circuit. 

R  may  be  separated  into  its  constituent  parts  by 
calling  Rg  the  resistance  of  the  generator,  Re  the  resist- 
ance of  the  conductors  including  any  extra  resistance 
inserted  to  regulate  the  current  and  Re  the  resistance 
of  the  electrolytic  cell  :  then  equation  (15)  takes  the  form 

E  =   C  (Rg  +  Re  -f  Re)  +  e      (16) 
and  solving  for  C  we  have 


The  effective  electromotive  force  in  the  circuit  is 
lower  than  the  electromotive  force  of  the  generator  by 
the  counter-electromotive  force  of  the  electrolytic  cell, 
so  the  current  is  equal  to  the  difference  of  these  electro- 
motive forces  divided  by  the  total  resistance.  For  any 
value  of  E  smaller  than  e,  no  current  will  flow  and  no 
metal  be  deposited  ;  so  this  counter-electromotive  force 
is  also  the  value  which  the  applied  electromotive  force 
must  exceed  in  order  to  produce  decomposition. 

This  fact  permits  the  deposition  of  a  single  metal 
from  solutions  containing  several.  For  instance  the  heat 
of  combination  per  equivalent  of  nickel  sulphate  in 
water  is  43475  calories  which  means  that  it  requires 

4v54/_o_  =  i  .89  volts  to  cause  its  decomposition  and 
23000 

deposit  nickel  :  to  deposit  copper  from  copper  sulphate 
requires  1.22  volts.  Therefore  if  we  apply  a  voltage 
above  1.22  and  below  1.89  to  a  cell  containing  a  mix- 
ture of  copper  and  nickel  sulphate  in  solution  only 
copper  will  be  deposited.  Given  a  solution  containing 


142 


OHM'S    LAW   FOR   ELECTROLYSIS. 


silver  and  copper  as  nitrates.  What  are  the  limits  of 
voltage  to  deposit  silver  and  not  copper  ?  The  heats  of 
combination  per  equivalent  of  these  salts  in  aqueous 
solution  are  respectfully  8390  and  26205  calories,  div- 
iding by  23000  we  get  the  counter-electromotive  force 
as  0.365  volt  for  silver  nitrate  and  1.14  volts  for  copper 
nitrate.  So  any  current  with  a  voltage  above  0.365  and 
below  1.14  will  deposit  only  silver  from  this  solution. 

When  all  of  the  first  metal  is  deposited  the  current 
will  fall  to  zero  unless  the  voltage  is  raised  above  the 
counter-electromotive  force  of  the  next  higher  metal  in 
solution. 

A  convenient  method  of  regulating  the  voltage  is  shown  by 
the  diagram. 

A  and  B  are  the  poles  of  the  source  of  current.  The  circuit 
is  closed  through  a  conductor  CD  whose  resistance  is  sufficient 
to  cut  down  the  current.  The  resistance  of  this  conductor  must 
of  course  depend  on  the  source  of  current,  which  may  be  either 

primary  batteries,  second- 
ary batteries,  dynamo  or 
JD  lightning  circuit.  The  elec- 
trolytic cell  is  connected  so 
that  one  pole  E  connects 
with  C  and  the  other  pole 
F  connects  with  the  mov- 
able point  or  slider  G  so 
that  any  point  of  the  con- 
ductor CD  can  be  selected 
and  any  voltage  from  zero 
(when  G  is  at  C)  to  the  full 
voltage  of  the  source  (when  G  is  at  D).  By  having  a  voltmeter 
V  connected  across  the  cell  and  starting  with  G  at  C  the 
proper  voltage  can  be  applied  by  pushing  G  towards  D  until  the 
voltmeter  indicates  the  value  desired. 

The  relation  existing  between  the  voltage  measured 


OHM'S   LAW   FOR    ELECTROLYSIS.  143 

across  an  electrolytic  cell  by  a  voltmeter  and  the  current 
passing  through  the  cell  is  given  by  equation  (16);  when 
Rg  and  Re  are  made  zero,  this  becomes 

E  =  CRc  +  e  (18) 

If  Re  be  negligible,  then  unless  C  is  very  great  E  = 
e,  or  the  voltage  across  the  cell  is  practically  a  constant, 
equal  to  the  counter-electromotive  force  of  the  cell  and 
independent  of  the  current  sent  through  it. 

If  either  Re  or  C  is  so  great  that  e  may  be  neglected 
then  the  voltage  is  proportional  to  the  current  flowing 
as  in  the  case  of  simple  conductor. 

In  most  cases,  however,  neither  can  be  neglected 
and  the  voltage  across  the  cell  depends  not  only  on  the 
counter-electromotive  force  but  also  on  the  resistance 
of,  and  the  current  flowing  through  the  electrolytic  cell. 
Let  us  consider  an  electrolytic  cell  containing  a  satura- 
ted solution  of  zinc  sulphate  with  platinum  electrodes  5 
centimeters  apart  whose  opposite  submerged  surfaces 
are  15  square  centimeters,  since  the  specific  resist- 
ance is  33.7  ohms  Re  in  this  case  will  be  (equation  (2)), 

33-7  X    —  =    1 1.23  ohms. 

The  counter-electromotive  force  of  zinc  sulphate  is 
2.31  volts  (see  tables)  if  the  current  passing  is  one-tenth 
of  an  ampere,  the  voltage  across  the  cell  will  be  found 
by  equation  (18) 

E  =  o.i  X   11-23  +    2.31    =   3.43  volts. 

If  the  current  is  raised  to  two-tenths  of  ampere  the 
voltage  will  not  be  double,  6.86,  but  only  4.56  volts. 
For 

E  =  0.2  X   11.23  +  2-3£    =  4-56  volts. 

If  the  cell  were  so  arranged   that  its  internal  resist- 


144  EFFECT     OF   DISSOLVING   ANODES. 

ance  were  only  one-tenth  of  an  ohm  for  small  currents 
the  voltage  across  its  poles  would  be  practically  constant : 
for  0.05  of  an  ampere,  2.315  volts;  foro.  loof  an  ampere; 
2.32  volts. 

Silver  nitrate  in  water  requires  only  0.365  of  a  volt  to 
effect  decomposition.     Therefore,  for  such  a  cell,  having 
a    resistance   of    100    ohms,    with    1/4   ampere  passing 
through  it,  the  voltage  across  the  poles  will  be 
E   =  V*  X   ioo  +  0.365   =    25.365. 

When  the  current  is  doubled  this  becomes 

E  =  %  X   ioo  +  0.365   =   50.365. 
or  practically  double. 

EFFECT    OF    DISSOLVING    ANODES. 

In  plating  processes  and  the  electrolytic  refining  of 
metals,  the  anode  of  the  bath  is  made  of  the  same  metal 
as  that  to  be  deposited,  so  that  as  the  current  passes  the 
anode  dissolves  at  the  same  rate  at  which  the  metal  is 
deposited  on  the  kathode ;  keeping  the  solution  at  the 
same  strength.  A  plating  or  refining  bath,  or  any  cell 
with  a  dissolving  anode  of  the  same  metal  as  in  the 
solution,  has  therefore  no  counter-electromotive  force, 
since  the  electromotive  forces  of  solution  and  decompo- 
sition are  equal  and  opposed  to  each  other. 

Take  the  familiar  example  of  a  cell  containing  copper 
sulphate,  the  anode  of  platinum,  the  kathode  of  copper; 
this  cell  has  a  counter-electromotive  force  of  1.22  volts. 
When  we  substitute  a  copper  anode  for  the  platinum, 
no  sulphuric  acid  is  formed  but  the  copper  forms  copper 
sulphate  with  the  SO4  and  the  anode  dissolves  at  the 
same  rate  as  the  copper  is  deposited.  The  formation  of 
copper  sulphate  ordinarily  generates  27,980  calories  per 
equivalent  in  grams,  but  here  the  energy  appears  as 


CURRENT    DENSITY.  145 

electricity  and  the  electromotive  force  produced  is  ex- 
actly equal  to  that  necessary  to  decompose  copper  sul- 
phate. A  bath  or  cell  of  this  kind  only  introduces  a 
certain  ohmic  resistance  into  the  circuit  (its  e  being 
zero)  and  for  it  formula  15,  16,  17,  1 8,  reduce  to  the 
simplest  form  of  Ohm's  law.  (3) 

CURRENT    DENSITY. 

The  character  of  the  deposit  depends  largely  on  the 
strength  of  current  used,  which  can  only  be  determined 
by  experiment.  When  this  has  been  found,  the  current 
is  often  described  as  having  a  certain  density,  that  is  a 
certain  number  of  amperes  per  unit  of  kathode  area. 

In  analytical  work  current  density  is  expressed  in 
amperes  per  100  square  centimeters  of  kathode  area; 
that  is  area  on  which  metal  is  deposited.  N.D.100  is  the 
abbreviation  used:  so  N.D.100  0.25  means  a  quarter  of 
an  ampere  for  each  100  square  centimeters  of  the  sub- 
merged area  of  the  kathode  on  which  metal  is  deposited. 
As  it  is  so  important  in  this  work  to  have  a  firm  adher- 
ent deposit  the  current  is  most  conveniently  described 
by  giving  the  voltage  and  current  density.  In  some  of 
the  older  books  on  the  subject  the  current  is  given  in 
cubic  centimeters  of  mixed  gases  per  minute,  in  order  to 
reduce  this  to  current  density  it  is  necessary  to  know 
that  one  ampere  equals  10.436  c.c.  of  oxyhydrogen  gas 
per  minute  measured  at  o°C  and  760  millimeters 
pressure. 


EXAMPLES, 

WITH    SOLUTION    OF    EACH. 

1.  How  many  amperes  are  flowing  in  a  circuit  which 
in  the  course  of  one  hour  and  1 5  minutes  has   deposited 
30. 1 86  grams  of  silver  ?  Ans.     6  amperes. 

Solution,  i  hour  and  15  minutes  =  4500  seconds  : 
one  ampere  by  definition  deposits  0.001118  gram  of 
silver  per  second.  Therefore  in  4500  seconds  it  will 
deposit  0.001118  X  4500,  or  5.031  grams.  Number  of 
amperes  flowing  equals  30.186  divided  by  5.031  or  6. 

2.  If  a  current  of  15  amperes  flows  for  one  hour  and 
20  minutes  how  many  coulombs  will  pass  ?     How  many 
ampere-hours  will  that  give  ? 

Ans.     72000  coulombs,  20  ampere-hours. 

Solution.      Substitute  in  equation  (i) 

Q  =  Ct  =  15  X  4800  =  72000  coulombs:  since  i  hr. 
and  20  minutes  =  4800  seconds.  72000  coulombs  -f- 
3600  =  20  ampere-hours. 

3.  If  the  specific  resistance  of  copper  is  0.000001629, 
what  is  the  resistance  of  a  wire  200  meters  long  having  a 
cross-section  of  2  square  millimeters  ? 

Ans.      1.629  ohms, 

Solution.      Substitute  in  equation  (2)  R  =  p  - 

a 

p  =  0.000001629,  1  =  20000  centimeters,  a  =  0.02  square 
centimeter. 

R  =0.000001629  -  —    -  =  1.629. 
0.02 

(146) 


EXAMPLES.  147 

4.  If   the    specific    resistance    of   German-silver     is 
0.0000209,  what  is   the  resistance  of  1 50  meters  of  wire 
having  a  cross-section  of  25  square  millimeters. 

Ans.      1.254  ohms. 

Solution.      Substitute    in    equation  (2)  R  =  p  - 

a 

P  =  0.0000209,   1  =  15000  centimeters,  a  =  0.25   square 
centimeter 

R  =  0.0000209   -^ — -.  =  0.0000209  X  60000  =  1.254 

5.  What  is  the  resistance  of  an  electric  light  carbon 
1 2   inches  long  by  ]/4  inch  in  diameter  when   its  specific 
resistance  is  0.07.      It  will  be  sufficient  to  call  2  l/2  centi- 
meters one  inch.  Ans.     7  ohms. 

Solution.      Substitute   in  equation  (2)  R  =  p  *— 

cL 

p  =  0.07,  1  =  12  X  2.5.=  30  centimeters,  a  =  nrz  = 

n  (y%  x  2.5)2  =  0.3  very  closely.  R  =  0.07  ^—    =    7  ohms 

practically. 

6.  What  is  the  resistance  of  a  saturated   solution  of 
sulphate  of  copper  placed   in  a  tube  10  c.m.   long  and 
having  a  cross-section  of  2  square  centimeters,  when  its 
specific  resistance  is  29.3.  Ans.      146.5  ohms. 

Solution.     Substitute  in  equation  (2)  R  =  p  — 

cl 

p  —  29.3,  1  =  10,  a  =  2.    R   =  29.3  _  =  146.5 

7.  What  is  the  total  resistance  in  a  circuit  composed 
of  a  voltaic  cell,  a  copper  wire,  an  iron  wire  and  a  piece 
of  German-silver  all  connected  in  series  when  the  separ- 
ate resistances  are  respectively  3,  2,    4,  6  ? 

Ans. .    15  ohms. 


148  EXAMPLES. 

Solution.  Total  resistance  of  resistances  connected 
in  series  equals  their  sum. 

8.  What   is   the   resistance    of  eleven    incandescent 
lamps  connected  in  parallel  if  each  has  a  resistance  of 
220  ohms  when  hot.  Ans.      20  ohms. 

Solution.  Resistance  of  equal  resistance  connected 
in  parallel  is  equal  to  resistance  of  one  divided  by  their 
number. 

9.  What  is  the  resistance  of  four  wires  of  respectively 
20,  5,  4,  and  2  ohms  connected  in  parallel. 

Ans.      One  ohm. 
Conductance  of  a  number  of  conductances  connected 

in  parallel  equals  their  sum.   — '   — '  — '  —  equals  .05   4- 

20     5      4      2 

.2  -f-  .25  -f-  -50  =  i.     Reciprocal  of  one  is  one. 

10.  If  6  amperes  are  flowing  through  7  ohms,  what  is 
the  voltage  measured  between  the,ends  of  this  resistance? 

Ans.     42  volts. 
Solution.     Ohm's  law,  E  =  CR  =  6  X  7  =  42. 

11.  If  a  voltage  of  17  is  applied  to  34  ohms  how 
many  amperes  will  pass?  Ans.     0.5  ampere. 

Solution.     C  =  — =  _/  =  0.5 
R      34 

1 2  What  is  the  resistance  of  a  conductor  that  allows 
%  of  an  ampere  to  flow  when  a  voltage  of  75  is  applied 
to  its  terminals?  Ans.  100  ohms. 

Solution.      From  Ohm's  Law  R  =  _  =    ^5    =IOO 

C       0.75 

13.  If  a  wire  of  2  ohms  resistance  is  immersed 
in  loo  grams  of  distilled  water  how  long  will  it  take 
%  of  an  ampere  to  raise  the  temperature  of  the  water 
from  4  to  6  degrees  centigrade,  assuming  no  loss  from 
radiation.  Ans.  12  minutes  21  seconds. 


EXAMPLES.  149 

Solution.     100  grams  of  water  raised  2°  =  200  calor- 
ies.    Substitute  in  formula  (4) 

V  -  C2Rt  X  0.24.  V  =  200,  C  =  3/4,  R  =  2. 

200  =  —  X  2  X  t  X  0.24  =  0.27  t.        t  =  741  seconds 
16 

or  12  minutes  21  seconds. 

14.  How  many  joules  would  be  expended  as  heat  by 
3  amperes  flowing  through  2  ohms  for  10  minutes. 

Ans.      10,800  joules. 

Solution.     Substitute  in  equation  (5)  W  =  C2Rt. 
C  =  3,  R  =  2,  t  =  600.  W  =  9  X  2  X  600  =  10800. 

15.  An    enclosed   arc  lamp  requires    5    amperes   to 
burn  properly.     When  placed  on   an    1 1 8   volt  circuit, 
how  many  watts  are  taken  ?  Ans.     590  watts. 

Solution.     Substitute  in  equation  (7) 
P  =  CE       C  =  5,  E  =  118.     P  =  5  X  118  =  590. 

1 6.  How  many  watts  do  25  incandescent  lamps  take 
connected  in  parallel  from  a  no  volt  circuit,  when  each 
lamp  draws  half  an  ampere  ?     How  many  kilowatts  and 
how  many  horse-power  does  this  make  ?      Ans.      1375 
watts,  1.375  kilowatts  and  1.84  horse-power. 

Solution.      25  lamps  at  Y2  ampere  apiece  gives   12.5 
amperes.      Substitute  in  equation  (7)     P  =   CE, 
C  =  12.5,  E  =  no,  P  =  12.5  X  no  =  1375. 
1000  watts  =  i  kilowatt.      1375  watts  =  1.375  kilowatts. 
746  watts  =    i  horse-power.      1375  watts  =  1.84  horse- 
power. 

1 7.  At  full  load  a  dynamo   delivers  600  amperes  at 
125  volts  what  is  its  output  in  watts  and  what  would  be 
its  rating  in  kilowatts  and  horse-power. 

Ans.     75000  watts,  75  kilowatts,  100  horse-power. 
Solution.      Substitute  in  equation  (7) 

P  —  CE.     C  =  600,  E  =  125,  P=  75000. 


150  EXAMPLES. 

75°°°  —  75  kilowatts,    i   kilowatt  =    i  1/3  horse-power, 
1000 
75  X  i  l/s  —  IO°  horse-power. 

1 8.  How  much  zinc  will  be  deposited  from  zinc  sul- 
phate by  3  amperes  in  10  minutes  ?    Ans.   0.6098  gram. 

Solution.      Substitute  in  equation  (8) 
w  =  0.00001036  ACt.  A  =  32.7,   C   =  3,  t  =  600. 
w  =  0.00001036  X  32-7  X  3   X  600  =  0.0003388  X 
1800  =  0.6098  gram. 

19.  How  much  lead  will  be  deposited  by  a  current 
of  1/2  an  ampere  passing  for  90  minutes  through  lead 
chloride.  Ans.     2.894  grams. 

Solution.      Substitute  in  equation  (8) 
w  =  0.00001036  ACt     A  =   103.5,   C  Yg,  t  =  5400. 
w  ==  0.00001036  X  103.5  X  l/s  X  5400 
=  0.001072  X   2700  =   2.894  grams. 

20.  How  much  chlorine  would  be  liberated  by  1/2  an 
ampere  passing  through  dilute  hydrochloric  acid  for    13 
hours.  Ans.     8.593  grams. 

Solution.     Substitute  in  equation  (9) 
w  =  0.0373  ACN.     A  =  35.45,  C=  V»  N  =  13. 
w  =  0.0373  X  35-45  X  l/2  X  13 
=  I.322X  6.5  =  8.593  grams. 

21    How  much  tin   will   be  deposited  by  5  amperes 
flowing  6  hours  through  a  solution  of  SnCl2? 

Ans.     66.6  grams. 
Solutiun.      Substitute  in  equation  (9) 
w  =  0.0373  ACN.     A  =  59.52,  C  =  5,  N  =  6 
w  =  0.0373  X  59-52  X  5  X  6 
w   =   2. 2 20    X  30  =  66.6  grams. 
22.    How  many  amperes  will  it  take  to  deposit    2^4 
pounds  of  copper  from  copper  sulphate  in  10  hours? 

Ans.     95.7  amperes. 


EXAMPLES.  151 

Solution.      Substitute  in  equation  (10) 

^        26.8  w  ,  ., 

C   =  .    w  =  1136  grams,     i  kilo  =  2.2  Ibs.  hence 

A.  IN 

2.5  Ibs  =  1.136  kilos  =  1  136  grams. 
A  =  31.8,  N  =,o.     C  = 


^  I  .  o    X    I  O 

23.  What  is  the  minimum  voltage  necessary  to  de- 
compose water  if  the  heat  of  combination  of  one  gram 
of  hydrogen  with  oxygen  is  34180  calories. 

Ans.     1.486  volts. 

Solution.      Substitute  in  equation  (14) 


e  =        H       =  =        86  volts 

23000        23000 

24.  If  zinc  bromide  requires  1.65  volts  to  decompose 
it,  what  is  the  heat  evolved  by  the  equivalent  weight  of 
zinc  in  grams  combining  with  bromine? 

Ans.   37950  calories. 

Solution.      From  equation  (14)  H  =  23000  e  hence 
H  =  23000  X  1-65  =  37950. 

25.  If  one  gram  of  hydrogen  combining  with  chlor- 
ine in  the  presence  of  water  evolves  39315  calories  what 
is  the  voltage  required  for  the  decomposition  of  hydro- 
chloric acid?  Ans.      1.71  volts. 

Solution.     Substitute  in  equation  (14) 

e  =  -H      =393*5    =   1.71  volts. 
23000       23000 

26.  It  is  desired  to  send  half  an  ampere  through  ten 
ohms  of  copper  sulphate.     What  resistance  must  be  put 
in  series  with  the  cell  if  the  current  is  to  be  drawn  from 
a  circuit  whose  voltage  is  1  18.     What  will  be  the  values 
of  this  extra  resistance  for  respectively   ^   and    2    am- 
peres ?  Ans.     223.56,  457.12,  and  48.39  ohms. 


152  EXAMPLES. 

Solution.     Neglecting  the  resistance  of  the  generator 
and  leads  equation  (16)  becomes  for  this  example: 
E=  C(R!+Rc)  +  e.  E  =  118,  C  =  ya,  Rc=  io,e  =  1.22 
118  =  y2  (R!  +  10)  +  1.22 
yg  R!  =  1 18  -  5 — 1.22  =  1 1 1.78      R!  =  223.56  ohms. 

if  c  =  y4.  us  =  y4(Ri+  io)+  1.22 

y4  Rj  =  1 18—  2.5  -  1.22  =  1 14.28  R,  =  457.  i  2  ohms 

If  C    =    2.      Il8   =.2(Ri+    10)  +    1.22 

2R!=ii8  —   20   -    1.22=96.78.      R!  =  48.39  ohms. 

27.  Two  batteries,  having  a  negligible  internal  resist- 
ance   and   a  voltage  of  two    each,  are  to   furnish    cur- 
rent to  be  sent  through  a  solution   of  zinc   sulphate    of 
2.7  ohms  resistance.     What  extra  resistance  must  be  in- 
serted to  limit  the  current  to  half  an  ampere  ?     How 
much  more  must  be  added  to  cut  the  current  down  to 
y4  of  an  ampere  ? 

Ans.     0.68  of  an  ohm,  3.38  ohms  additional. 
Solution.     Substitute  in  E  =  C^  +  Rc)+  e 
E  =  4.,  C  =  ya,  Rc  =  2.7,  e  =  2.31    (see  tables) 
4  =  %  (R,  +  2.7)  +  2.31-     %  R!  =  4  -  1-35  -  2.31. 

yg  Rj  =  0.34.     R!  =  0.68  of  an  ohm. 

If  C  =  %  4  =  'A  (Ri+  2.7)  +  2.31.  V4  R!  =  4-0.675 
-  2  31  =  1.015.  R!  =  4.06.  4.06—  0.68  =  3.38  ohms 
additional. 

28.  What  current  will  pass  through  a  tube  of  acidu- 
lated water,  having  a  resistance  of  25  ohms,   between 
platinum  electrodes  when  connected  to  a  1 18  volt  circuit 
in  series  with  6  lamps  in  parallel  assuming  each  lamp  to 
be  240  ohms  resistance.     What  will  be  the  value  of  the 
current  when  half  the  lamps  are  unscrewed  ? 

Ans.      1.79  and  1. 1 1  amperes. 
Solution.      Substitute  in  equation  (17) 


EXAMPLES.  153 

E  =  1  1  8,  e  =  1.486    example   (23)    Rg  =  o,    neglected, 


=_  =  I>?9 

40  +  25  65 

For  3  lamps  R!  =  ^o  =  80.    C  -  "8-i. 


80+25  105 

=  1.11  amperes 

29.  What  is  the  voltage  across  an  electrolytic  cell  of 
silver  nitrate  having  a  resistance  of  40  ohms  when  half 
an  ampere  is  passing.  Ans.     20.365  volts. 

Solution.  Substitute  in  equation  (18)  E  =  CRC+  e. 
C  =  %  Rc  =  40,  e  =  0.365  (see  tables)  E  =  l/2X  40  -f 
0.365  =  20.365  volts. 

30.  What  is  the  voltage  across  an  electrolytic  cell  of 
zinc  bromide  of  2  ohms  resistance  when   the  circuit  is 
closed   by  inserting   \l/z  ohms   from   a  storage    battery 
giving  4  volts.  Ans.     3  volts  practically. 

Solution.      First  find  C  from  equation  (17) 

C  =  -      ~e      E  =  4,  e  =  1.65    (example  24)   R,  =  1.5 
KI  -f  Kc 

Rc=    2. 

C  =  3~  T'65  =   M5  =  067  practically 

i-S+2          3.5 

Now  substitute  in  equation  (18)  E  =  0.67  X  2  +  1.65 
=  1.34  +  1.65  =  2.99  volts,  practically  3. 


TABLES. 


TABLES   OF    WEIGHTS. 

TABLES  OF  WEIGHTS. 


157 


METRIC  SYSTEM. 


Milligram 

Gram 

Gram 

Gram 

Kilogram 

Kilogram 


0.015438395  grains. 
15.438395  grains. 
0.03527  ounces  avoirdupois. 
0.03216  ounces  troy. 
2.205486  pounds  avoirdupois. 
2.6803  pounds  troy. 


AVOIRDUPOIS. 

Long  ton       =       2240  pounds       =     1015.649  kilograms. 
Short  ton       =       2000  pounds       =       906. 8296  kilograms. 

Pound    =  16  ounces    =    7000  grains    =    453.4148  grams. 
Ounce  437.5  grains  =      28.3375  grams. 

Grain  =          64.773  milligrams    =        0.064773  grams. 


TROY. 

Pound  =  12  ounces  =  5760  grains  =  373.096  grams. 
Ounce  =  20  pennyweights  =  480  grains  =  31.0913  grams. 
Pennyweight  =  24  grains  =  "•  x*-*^  ™-«™ 

Grain      =    64.773  milligrams 


1.55457  grams. 
0.064773  grams. 


TROY   (Pharmacy.) 


Ounce  =  8  drams  =  480  grains  •=  31.0913  grams. 
Dram  =  3  scruples  =  60  grains  =  3.8869  grams. 
Scruple  =  20  grains  =  1.29546  grams. 


158  TABLES    OF   MEASURES. 

TABLES  OF  MEASURES. 

LENGTH. 

Millimeter  0.03937  inch- 
Centimeter  =  0.393703  inch. 
Decimeter  3.93708  inches. 
Meter  39.3708  inches. 
Meter  3. 2809  feet. 
Meter  1.093633  yards. 
Inch  2.539954  centimeters. 
Foot  —  12  inches  3.047945  decimeters. 
Yard  =  3  feet  0.914383  meter. 
Mile  =  1700  yards  5280  feet. 
Mile  =  1.609315  kilometers  =  1609.315  meters, 

SURFACE. 

Square  millimeter  0.00155  square  inches, 

centimeter  0.155086       "  " 

decimeter  15.5086 

"       decimeter  0.10769       "       foot. 

"       meter  =  1550.86        "       inches. 

"       meter  =  10.7698  square  feet  =  1.196     "       yards. 

VOLUME. 

Gallon  U.  S.                            =  231  cubic  inches. 

Gallon     "  3. 78537  liters. 

Quart      "  0.94634  liter. 

Pint         "  0.47317  liter. 

Liter  =  2.1134  pints  U.  S.  =  1.0567  quarts  U.  S. 
=  0.2641 7  gallon  U.  S. 

Cubic  meter  =  1.308  cubic  yards    =    35.3171  cubic  feet. 

An  imperial  gallon,  English  =  277. 274  cubic  inches 
=  4. 543458  liters. 


ATOMIC   WEIGHTS. 

ATOMIC  WEIGHTS. 


159 


Reported   by  the  Committee  on  Atomic  Weights  to  the  American 
Chemical  Society.* 


H  =  l 

O=16 

H  =  1 

O=16 

Aluminium 

26.91 

27.11 

Neodymium 

142.52 

143.60 

Antimony 

119.52 

120.43 

Nickel 

58.24 

58.69 

Arsenic 

74.44 

75.01 

Nitrogen 

13.93 

14'.  04 

Barium 

136.39 

137.43 

Osmium 

189.55 

190.99 

Bismuth 

206.54 

208.11 

Oxygen 

15.88 

16.00 

Boron 

10.86 

10.95 

Palladium 

105.56 

106.36 

Bromine 

79.34 

79.95 

Phosphorus 

30.79 

31.02 

Cadmium 

111  54 

112.38 

Platinum 

193.41 

194.89 

Caesium 

131.89 

132.89 

Potassium 

38.82 

39.11 

Calcium 

39.76 

40.07 

Praseodymium 

139.41 

140.46 

Carbon 

11.91 

12.00 

Rhodium 

102.23 

103.01 

Cerium 

138.30 

139.35 

Rubidium 

84.78 

85  43 

Chlorine 

35.18 

35  45 

Ruthenium 

100.91 

101.68 

Chromium 

51.74 

52.14 

Samarium 

149.13 

150.26 

Cobalt 

58.55 

58.99 

Scandium 

43.78 

44.12 

Columbium 

93.02 

93.73 

Selenium 

78.58 

79.17 

Copper 

63.12 

63.60 

Silicon 

28.18 

28.40 

Erbium 

165.06 

166.32 

Silver 

107.11 

107.92 

Fluorine 

18.91 

19  06 

Sodium 

22.88 

23.05 

Gadolinium 

155.57 

156.76 

Strontium 

86.95 

87.61 

Gallium 

69.38 

69.91 

Sulphur 

31.83 

32.07 

Germanium 

71.93 

72.48 

Tantalum 

181.45 

182.84 

Glucinum 

9.01 

9.08 

Tellurium 

126.52 

127.49 

Gold 

195.74 

197.23 

Terbium 

158.80 

160.00 

Hydrogen 

1.00 

1.008 

Thallium 

202.61 

204  15 

Indium 

112.99 

113.85 

Thorium 

230.87 

232.63 

Iodine 

125.89 

126.85 

Thulium 

169.40 

170:70 

Iridium 

191.66 

193.12 

Tin 

118.15 

119.05 

Iron 

55.60 

56.02 

Titanium 

47.79 

48.15 

Lanthanum 

137.59 

138.64 

Tungsten 

183.43 

184.83 

Lead 

205.36 

206.92 

Uranium 

237.77 

239.59 

Lithium 

6.97 

7.03 

Vanadium 

50.99 

51.38 

Magnesium 

24.10 

24.28 

Ytterbium 

171.88 

173.19 

Manganese 

54.57 

54.99 

Yttrium 

88.35 

89.02 

Mercury 

198.49 

200.00 

Zinc 

64.91 

65.41 

Molybdenum 

95.26 

95.99 

Zirconium 

89.72 

99.40 

•J.  Am    Chem.  Soc.   Feb.  1899. 


160 


TABLE    OF    FACTORS. 

TABLE  OF   FACTORS. 


Required. 

Factor. 

Logarithm. 

A1203. 

Al. 

0.53042 

T.  72462 

A1PO4. 

Al. 

0.22200 

T.  34631 

A12O3. 

0.41849 

T.62168 

Sb204. 

Sb. 

0.79010 

T  89766 

Sb2S3. 

Sb. 

0.71457 

T.  85404 

As2S3. 

As. 

0.00922 

T.  78477 

Mg2As207. 

As. 

0.48300 

T.68395 

Ag3AsO4. 

As. 

0.16209 

T.  209  76 

BaSO4. 

BaO, 

0.65709 

T.  81762 

S03. 

0.34291 

T.  53518 

s. 

0.13734        f.  13781 

Bi203. 

Bi. 

0.89660    >   T.95260 

CdS. 

Cd. 

0.77798       T.  89097 

CaC03. 

CO2. 

0.43969        T.64315 

CaO. 

0.56037    ;   T.  74847 

CaF2. 

F. 

0.48753    •   T.  68800 

CaSO4. 

CaO. 

0.41186        T.61475 

CaCO3. 

0.73505       T.  86632 

C02. 

C. 

0.27272 

1.43572 

Cr203. 

Cr. 

0.68479 

T.  83556 

3K2SO4,2CoSO4. 

Co. 

0.14163 

T.15115 

CuO. 

Cu. 

0.79900 

T.  90254 

Cu2S. 

Cu. 

0.79864 

T.  90235 

Fe2O3. 

Fe. 

0.70008 

T.  84514 

Fe. 

FeaOa. 

1.42842 

0.15486 

FeO. 

1.28561 

0.10911 

Fe. 

Fe304. 

1.38082 

0.14014 

PbCrO4. 

Pb. 

0.64050 

T.  80652 

PbSO4. 

Pb. 

0.68293 

T.  83437 

Li3P04. 

Li2O.  ' 

0.38834 

T.  589  21 

LiCl. 

1.09758 

0.040436 

Mg2P2O7. 

P. 

0.27870 

T.  44514 

P205. 

0.63809 

T.  80488 

MgO. 

0.36190 

T.  55859 

TABLE   OF    FACTORS. 

TABLE  OF  FACTORS. 


161 


Required. 

Factor. 

Logarithm. 

MgC03. 

0.75722 

T.  87922 

Mn3O4. 

Mn. 

0.72049 

T.  85763 

Mn2P207. 

Mn. 

0.38723 

T.  58796 

MnS. 

Mn. 

0.63163 

T.  80046 

MnSO4. 

Mn. 

0.36403 

T.56113 

HgS. 

Hg- 

0.86181       T.  93541 

NiO. 

Ni. 

0.78578       T.  89530 

(NH4)2PtCl6. 

Pt. 

0.43920       T.  64267 

N 

0.06328       2.80127 

NH3. 

0.07691       2.88599 

NH4C1. 

0.24123       T.  38244 

Pt  from(NH4)2PtCl6. 

N. 

0.14408 

1.15861 

NH3. 

0.17511 

T.  24332 

NH4C1. 

0.54925 

T.  73977 

K2PtCl6. 

KC1. 

0.30695 

T.  48707 

K20. 

0.19394 

T.  28768 

KC1. 

K2O. 

0.63184 

T.8oo6i--A 

K2SO4. 

K2O. 

0.54059 

T.  73287 

K. 

0.44872 

T.  65205 

SiO2. 

Si. 

0.47020 

T.  6  7228 

AgBr. 

Br. 

0.42556 

T.  62896 

Agl. 

I. 

0.54031 

T.  73265 

AgCl. 

Cl. 

0.24726 

T.39316 

Ag- 

0.75274 

T.  87664 

NaCl. 

Na20. 

0  53077 

T.  72491 

Na2SO4. 

Na2O. 

0.43680 

T.  64028 

Na. 

0.32426 

T.  51089 

SrS04. 

SrO. 

0.56408 

T.  751  34 

SnO2. 

Sn. 

0.78815 

T.  89661 

TiO2. 

Ti. 

0.60075 

1.77869 

W03. 

W. 

0.79384 

T.  89973 

ZnO. 

Zn. 

0.80346 

T.  90497 

Zn2P2O7. 

Zn. 

0.42911 

T.  63257 

ZnNH4PO4. 

Zn. 

0.36644 

T.  56400 

In  most  cases  the  factor  to  four  places  is  sufficient. 


162 


SPECIFIC    HEATS. 

SPECIFIC  HEATS. 


Lithium, 0.941 

Carbon, 0.463 

Sodium, 0.273 

Magnesium, 0.245 

Aluminum, 0.225 

Silicon, 0.203 

Phosphorus, 0.202 

Sulphur, 0.178 

Potassium, 0.166 

Calcium, 0.169 

Titanium, 0.148 

Manganese, 0.122 

Iron, 0.112 

Nickel, 0.108 

Cobalt, 0.107 

Chromium, 0.100 


Copper, 0.0950 

Zinc, 0.0935 

Arsenic, 0.0830 

Molybdenum, 0.0659 

Silver 0.0570 

Cadmium, 0.0567 

Tin, 0.0559 

Antimony, 0.0523 

Tungsten, 0.0350 

Mercury, 0.0328 

Platinum, 0.0324 

Gold, 0.0324 

Iridium, 0.0323 

Lead, 0.0307 

Bismuth, 0.0305 

Uranium, 0.0276 


Thermometers. 

Three  scales  are  now  in  general  use.     These  are  : 

1.  Centigrade — C.     Water  freezes  at    0°,  boils  at  100°. 

2.  Fahrenheit— F.  "          32°,        "        212°. 

3.  Reaumur— R.  "  "  0°,        "          80°. 

5(F.°-32°)  * 


To  CONVERT — F.  to  C. 
C.  to  F. 
RtoF. 


9 
+32C 


"  Formulae. 


COMPARISON  OF  CENTIGRADE  AND  FAHRENHEIT  DEGREES.  163* 

COMPARISON  OF  CENTIGRADE  AND  FAHRENHEIT  DEGREES. 


C      F° 

C° 

F° 

c° 

F° 

c° 

F- 

•  c° 

F° 

—40 

—40 

22 

71.6 

48 

118.4 

74 

165.2 

100 

212 

—30 

—22 

23 

73.4 

4(J   120.2 

75 

167 

150 

302 

—20 

-  4 

24 

75.2 

50   122 

76 

168.8 

200 

392 

—10 

14 

25 

77 

51   123.8 

77 

170.6 

250 

482 

0 

32 

26 

78.8 

52   125.6 

78 

172.4 

300 

572 

1 

33.8 

27 

80.6 

53   127.4 

79 

174.2 

400 

752 

2 

35.6 

28 

82.4 

54   129.2 

80 

176 

500 

932 

3 

37.4 

29 

84.2 

55 

131 

81 

177.8 

600 

1112 

4 

39.2 

30 

86 

56 

132.8 

82 

179.6 

700 

1292 

5 

41 

31 

87.8 

57 

134.6 

83 

181.4 

800 

1472 

6 

42.8 

32 

89.6 

58 

136.4 

84 

183.2 

900 

1652 

7 

44.6 

33 

91.4 

59 

138.2 

85 

185 

1000 

1832 

8 

46.4 

34 

93.2 

60 

140 

86 

186.8 

1100 

2012 

9 

48.2 

35 

95 

61 

141.8 

87 

188.6 

1200 

2192 

10 

50 

36 

96.8 

62 

143.6 

88 

190.4 

1300 

2372 

11 

51.8 

37 

98.6 

63 

145.4 

89 

192.2 

1400 

2552 

12 

53.6 

38 

100.4 

64 

147.2 

90 

194 

1500 

2732 

13 

55.4 

39 

102.2 

65 

149 

91 

195.8 

1600 

2912 

14 

57.2 

40 

104 

66 

150.8 

92 

197.6 

1700 

3092 

15 

59 

41 

105.8 

67 

152.6 

93 

199.4 

1800 

3272 

16 

60.8 

42 

107.6 

68 

154.4 

94 

201.2 

1900 

3452 

17 

62.6 

43 

109.4 

69 

156.2 

95 

203 

2000 

3632 

18 

64.4 

44 

111.2 

70 

158 

96 

204.8 

2500 

4532 

19 

66.2 

45 

113 

71 

159.8 

97 

206.6 

3000 

5432 

20 

68 

46 

114.8 

72 

161.6 

98 

208.4 

31 

69.8 

47 

116.6 

73 

163.4 

99 

210.2 

164      TABLE  OF  VALUES  OF  NORMAL  SOLUTIONS. 

TABLE  OF   VALUES  OF   NORMAL   SOLUTIONS. 


Name. 

Formula. 

Molecular 
Weight. 

Weight  in 
Grams 
Per  Liter. 

Value  in 
Grams   of 
1.  C.  C. 

Sodium  Oxide 

Xa20 

62.1 

31.05 

0.03105 

Sodium  Hydroxide 

NaOH 

40.06 

40.06 

0.04006 

Sodium  Carbonate 

Na2CO3 

106.10 

53.05 

0  05305 

Sodium  Bicarbonate 

NaHCO3 

84.06 

84.06 

0.08406 

Potassium  Oxide 

K2O 

94.22       47.11 

0.04711 

Potassium  Hydroxide 

KOH 

56.12 

56.12 

0.05612 

Potassium  Carbonate 

K2C03 

138.22 

69.11        0.06911 

Potassium  Bicarbonate 

KHCO3 

100.12 

100.12 

0.10012 

Ammonia 

NH3 

17.06 

17.06 

0.01706 

Ammonium  Chloride 

NH4C1 

53.52 

53.52 

0.05352 

Calcium  Oxide 

CaO 

56.07 

28.035 

0.028035 

Calcium  Carbonate 

CaCO3 

100.07 

50.035 

0.050035 

Calcium  Sulphate 

CaSO4 

136.14 

68.07 

0.06807 

Barium  Hydroxide 

Ba(OH)2 

171.45 

85.725 

0.085725 

Barium  Carbonate 

BaCO3 

197.44      98.72 

0.09872 

Barium  Chloride 

BaCl2 

208.33    \  104.165 

0.104165 

Magnesium  Oxide 

MgO 

40.28        20.14 

0.02014 

Sodium  Sulphate 

Na2SO4 

142.17 

71.085 

0.071085 

Potassium  Sulphate 

K2SO4 

174.29 

87.145 

0.087145 

Sulphuric  Acid 

H2S04 

98.09 

49.045 

0.049045 

Nitric  Acid 

HNO3 

63.05 

63.05 

0.06305 

Hydrochloric  Acid. 

HC1 

36.46 

36.46 

0  03646 

Oxalic  Acid 

H2C204 

90.02 

45.01 

0.04501 

This  and  the  following  table  are  based  on  Oxygen    16.  and 
.the  atomic  weights  given  on  page  159. 


VALUES  OF  TENTH  NORMAL  SOLUTIONS. 


165 


VALUES  OF  TENTH  NORMAL  SOLUTIONS. 


Name 

Formula. 

Atomic  or 
Molecular 
Weight. 

Weight  in 
Grams 
per  Liter. 

Value  in 
Grms.  of 
One  C.C. 

Potassium  Permanganate  .... 

KMn04 
Fe 

158.1 
56  02 

3.162 
5  602 

0.003162 

0  005602 

Fe»O 

160  04 

8  002 

0  008002 

Ferrous  Oxide  
Magnetic  Oxide  

FeO 

Fe3O4 

72.02 
232.06 

7.202 
7.735 

0.007202 
0  007735 

Ferrous  Sulphate  (Cryst)  
Ferrous  Ammonium  Sulphate 
Oxalic  Acid  (Cryst)  
Ammonium  Oxalate  (Cryst)  .  . 
Calcium  Oxide  
Calcium  Carbonate  
Calcium  Sulphate  (Cryst)  
Manganese  (Volhard)  
Manganese  (Ford-Williams).  . 

FeSO4.  7H2O 
FeSO4.  (NH4)2SO4.  6H2O 
HaCa04.  2HaO 
(NH4)aC204,  HaO 
CaO 
CaCO, 
CaSO4.  2H2O 
Mn. 
Mn. 
p 

278.20 
392.40 
126.05 
142.16 
56.07 
100.07 
172.17 
54.99 
54.99 
31  02 

27.820 
39.24 
6.3025 
7.108 
2.8035 
5.0035 
8.6085 
1.65 
2.75 
0  08616 

0.027320 
0.039240 
0.0063025 
0.007108 
0.0028035 
0.0050035 
0.0086085 
0.00165 
0.00275 
0  0000861  ( 

Phosphoric  Oxide  (Noyes).  .  . 
Tin  from  SnCl2  
Hydrogen  Sulphide  
Hydrogen  Peroxide  
Iodine  

P,OB 

Sn 
H2S 
HaOa 
I 

142.04 
119.05 
34.09 
34.02 

126.85 

0.1973 
5.9525 
1.7043 
1.701 
12  685 

0.0001973 
0.0059525 
0.0017043 
0.001701 
0  012685 

Copper  

Cu 

Sb 

63.6 

120  43 

6.36 
6  021 

0.00636 
0  006021 

Sulphurous  Acid  .  ,  

HaS08 
SO, 

82.09 
64.07 

4.1045 
3  2035 

0.004104 
0  003203 

Sodium  Thiosulphate 

NaaS2O3 

158.24 

15  824 

0  015824 

Sodium  Thiosulphate  (Cryst). 
Chromium  
Chromic  Oxide  

Na,Sa03.  5H,0 
Cr 
CraOs 

248.32 
52.14 
152.28 

24.832 
1.738 
2  538 

0.024832 
0.001738 
0.002538 

Potassium  Dichromate  
Chlorine 

K.Cr.O, 
Cl 

294.5 
35.45 

4.9083 
3  545 

0.004908 
0  003545 

Potassium  Chloride  
Sodium  Chloride  
Ammonium  Chloride  
Silver  
Silver  Nitrate  

KC1 
NaCl 
NH4C1 
Ag 
AgNO3 

74.56 
58.5 
53.52 
107.92 
16996 

7.456 
5.85 
5.352 
10.792 
16  996 

0.007456 
0.00585 
0.00535 
0.01079 
0.0170 

166 


DEGREES   BAUME    AND   SPECIFIC   GRAVITY. 


Table  of  Comparison — Degrees  of  the  Baume  Hydrometer  and 
Specific  Gravity.  * 

TABLE   FOR  LIQUIDS  HEAVIER  THAN  WATER. 


Degrees 
Baume'. 

Specific 
Gravity. 

Degrees 
Baume 

Specific 
Gravity. 

Degrees 
Baume'. 

Specific 
Gravity. 

0 

1.000 

26 

1.206 

52 

1.520 

1 

1.007 

27 

1.216 

53 

1.535 

2 

1.013 

28 

.226 

54 

1.551 

3 

1.020 

29 

.236 

55 

1.567 

4 

.027 

30 

.246 

56 

1.583 

5 

.034 

31 

.256 

57 

1.600 

6 

.041 

32 

.267 

58 

.617 

7 

.048 

33 

.277 

59 

.634 

8 

.056 

34 

.288 

60 

.652 

9 

.063 

35 

.299 

61 

.670 

10 

.070 

36 

.310 

62 

.689 

11 

.078 

37 

1.322 

63 

.708 

12 

1.086 

38 

.333 

64 

.797 

13 

1.094 

39 

.345 

65 

.747 

14 

1.101 

40 

.357 

66 

.767 

15 

1.109 

41 

1.369 

67 

1.788 

16 

1.118 

42 

.382 

68 

1.809 

17 

1.126 

43 

1.395 

69 

1.831 

18 

1.134 

44 

.407 

70 

1.854 

19 

1.143 

45 

1.421 

71 

1.877 

20 

1.152 

46 

1.434 

72 

1.900 

21 

1.160 

47 

1.448 

73 

1.924 

22 

1.169 

48 

1.462 

74 

1.949 

23 

1.178 

49 

1.476 

75 

1.974 

24 

1.188 

50 

1.490 

76 

2.000 

25 

1.197 

51 

1.505 

TABLE   FOR   LIQUIDS  LIGHTER  THAN  WATER. 


Degrees 
Baume". 

Specific 
Gravity. 

Degrees 

U.IUITU-. 

Specific 
Gravity. 

Degrees 
Baumd. 

Specific 
Gravity. 

10 

1.000 

27 

.896 

44 

.811 

11 

.993 

28 

.890 

45 

.807 

19 

.986 

29 

.885 

46 

.802 

13 

.980 

30 

.880 

47 

.798 

14 

.973 

31 

.874 

48 

.794 

15 

.967 

32 

.869 

49 

.789 

16 

.960 

33 

.864 

50 

.785 

17 

.954 

34 

.859 

51 

.781 

18 

.948 

35 

.854 

52 

.777 

19 

.942 

36 

.849 

53 

.773 

20 

.936 

37 

.844 

54 

.768 

21 

.930 

.    3S 

.839 

55 

.764 

22 

.924 

39 

.834 

56 

.760 

23 

.918 

40 

.830 

57 

.757 

24 

913 

41 

.825 

58 

.753 

25 

.907 

42 

.820 

59 

.749 

26 

.901 

43 

.816 

60 

.745 

*  Crooke's  Select  Methods,  p. 


VOLUME  AND  WEIGHT  OF  WATER.  167 

VOLUME  AND  WEIGHT  OF  WATER  FROM  0°C  TO  *1°C.  # 


Tempera 
ture. 


0° 

1 

2 
3 

4 

5 

6 

7 

8 

9 

10 
11 
12 
13 
14 
15 


Volume  of  one 

Weight  of  one 

Tempera- 
ture. 

Volume  of  one 

Weight  of  one 

gram 

grair 

1.000126 

0.999874 

16° 

1.001025 

0.998976 

1.000070 

0.999930 

17 

1.001193 

0.998808 

1.000030 

0.999970 

18 

1.001373 

0.998629 

1.000007 

0.999993 

19 

1.0015(54 

0  998438 

1.000000 

1.000000 

20 

1.001768 

0.998235 

1.000008 

0.999992 

21 

1.001981 

0.998023 

1.000031 

0.999969 

22 

1.002204 

0.997801 

1.000069 

0.999931 

23 

1.002438 

0.997568 

1.000122 

0.999878 

24 

1.002681 

0.997326 

1.000188 

0.999812 

25 

1.002935 

0.997073 

1.000269 

0.999731 

26 

1.003199 

0.996811 

1.000363 

0.999637 

27 

1.003472 

0.996540 

1.000470 

0.999530 

28 

1.003788 

0.996226 

1.000590 

0.999410 

29 

1.004045 

0.995971 

1.000722 

0.999278 

30 

1.004346 

0.995673 

1.000867 

0.999134 

31    . 

1.004656 

0.995365 

For  the  most  recent  results  on  the  expansion  of  water,  which 
do  not  differ  materially  from  the  preceding,  see  Annalen  d. 
Physik  u.  Chemie  [N.  F.]  60,  340,  1897. 


Wied.  Ann.  47.40x5.  1892. 


168 


DENSITIES   AND   WEIGHTS   OF   GASES. 


Theoretical  Densities  of  Gases  Referred  to  Air  and  Weights  of  One 
Liter  in  Grams  at  0°C  and  760  m.  m.  Pressure. 


Name. 

Formula. 

Density. 

Weight  of 
One  Liter.* 

Acetylene,     .... 

C2H2 

0.89820 

1.16143 

Allylene,  

C3H4 

1.38194 

1.78692 

Ammonia,      .... 

NH3 

0.58890 

0.76148 

Arsine,      

AsH3 

2.69728 

3.48772 

Bromine,  

Br2 

5.52271 

7.14115 

Carbon  Dioxide, 

C02 

1.51968 

1.96503 

Carbon  Monoxide,  . 

CO 

0.96709 

1.25050 

Carbon  Oxysulphide,  . 

COS 

2.07483 

2.68287 

Chlorine,  

C12 

2.44921 

3.16696 

Cyanogen,      .... 

(CN), 

1.79907 

2.32630 

Ethane,     

C2H6 

1.03667 

1.34047 

Ethylene,       .... 

C2H4 

0.96744 

1.25095 

Hydriodic  Acid,       .     . 

HI 

4.41570 

5.70972 

Hydrobromic  Acid, 

HBr 

2.79597 

3.61534 

Hydrochloric  Acid, 

HC1 

1.25922 

1.62824 

Hydrofluoric  Acid, 

HF 

0.69178 

0.89451 

Hydrogen,      .     .      .     . 

H2 

0.069234 

0.089523 

Hydrogen  Sulphide,     . 

H2S 

1.17697 

1.52189 

Methane 

CH*     • 

0*^907 

0  71  ^02 

Nitric  Oxide, 

VxXT.4 

NO. 

.  OO*Ct7  i 

1.03764 

\J,  t  -L  tJU/v 

1.34172 

Nitrogen,  

N2 

0.97010 

1.25440 

Nitrous  Oxide,    .     .     . 

N20 

1.52269 

1.96892 

Oxygen,     

02 

1.10521 

1.42908 

Phosphine,     .... 

PH3 

1.17552 

1.52001 

Silicon  Tetrafluoride,  . 

SiF4 

3.60469 

4.66105 

Sulphur  Dioxide,     . 

SO2 

2.21295 

2.86146 

Water  Vapor,      .     .     . 

H20 

0.62182 

0.80405 

Air,      

1.00000 

1.29305  ' 

Taken  from  Hempel's  Gas  Analysis,  Dennis. 

*  These  numbers  are  also  the    weight  of  a  cubic  meter  in. 
Kilograms. 


VAPOR    PRESSURE   OF    WATER. 


169 


VAPOR  PRESSURE  OF  WATER  FROM  O°  C  TO  100°  C 
IN  MILLIMETERS  OF  MERCURY.* 


Temp. 

Pressure. 

Temp. 

Pressure. 

Temp. 

Pressure. 

Temp. 

Pressure. 

0 

4.569 

25 

23.517 

50 

91.98 

75 

288.76 

1 

4.909 

26 

24.956 

51 

96.66 

76 

301.09 

2 

5.272 

27 

26.471 

52 

101.55 

77 

313.85 

3 

5.658 

28 

28.065 

53 

106.65 

78 

327.05 

4 

6.069 

29 

29.744 

54 

111.97 

79 

340.73 

5 

6.507 

30 

31.51 

55 

117.52 

80 

354.87 

6 

6.972 

31 

33.37 

56 

123.29 

81 

369.51 

7 

7.466 

32 

35.32 

57 

129.31 

82 

384.64 

8 

7.991 

33 

37.37 

58 

135.58 

83 

400.29 

9 

8.548 

34 

39.52 

59 

142.10 

84 

416.47 

10 

9.140 

35 

41.78 

60 

148.88 

85 

433.19 

11 

9.767 

36 

44.16 

61 

155.95 

86 

450.47 

12 

10.432 

37 

46.65 

62 

163.29 

87 

468.32 

13 

11.137 

38 

49.26 

63 

170.02 

88 

486.76 

14 

11.884 

39 

52.00 

64 

178.86 

89 

505.81 

15 

12.674 

40 

5487 

65 

187.10 

90 

525.47 

16 

13.510 

41 

57  87 

66 

195.67 

91 

545.77 

17 

14.395 

42 

61.02 

67 

204.56 

92 

566.71 

18 

15.330 

43 

64.31 

68 

213.79 

93 

588.83 

19 

16.319 

44 

67.76 

69 

223.37 

94 

610.64 

20 

17.363 

45 

71.36 

70 

233.31 

95 

633.66 

21 

18.466 

46 

75.13 

71 

243.62 

'  96 

657.40 

22 

19.630 

47 

79.07 

72 

254.30 

97 

681.88 

23 

20.858 

48 

83.19 

73 

265.38 

98 

707.13 

24 

22.152 

49 

87  49 

74 

276.87 

99 

733.16 

*  Taken  from  Ostwald's  Manual  of  Phvsico-Chemical  Measurements. 


170 


SPECIFIC    GRAVITY    OF   ALCOHOL. 


TABLE   OF  SPECIFIC  GRAVITY  AND   PERCENTAGE  OF 
ALCOHOL   (ETHYL).* 


Per  Cent, 
by 
Volume. 

Per  Cent. 

by 
Weight. 

Specific 
Gravity  at 

I5.56°C. 

Per  Cent, 
by 
Volume. 

PerCent 

by 
Weight. 

Specific 
Gravity  at 

iS-56°C. 

0 

0.00 

1.0000 

51 

43.47 

.9315 

1 

0.80 

.9976 

52 

44.42 

.9295 

2 

1.60 

.9961 

53 

45.36 

.9275 

3 

2.40 

.9947 

54 

46.32                  .9254 

4 

3.20 

.9933 

55 

47.29 

.9234 

5 

4.00 

.9919 

56 

48.26 

.9213 

6 

4.81 

.9906 

57 

49.23 

.9192 

7 

5.62 

.9893 

58 

50.21 

.9170 

8 

6.43 

.9881 

59 

51.20 

.9148 

9 

7.24 

.9869 

60 

52.20 

.9126 

10 

8.05 

.9857 

61 

53.20 

.9104 

11 

8.87 

.9845 

62 

54.21 

.9082 

12 

9.69 

.9834 

63 

55.21 

.9059 

13 

10.51 

.9823 

64 

56.22 

.9036 

14 

11.33 

.9812 

65 

57.24 

.9013 

15 

12.15 

.9802 

66 

58.27 

.8989 

16 

12.98 

9791 

67 

59.32 

.8965 

17 

13.80 

.9781 

68 

60.38 

.8941 

18 

14.63 

.9771 

69 

61.42 

.8917 

19 

15.46 

.9761 

70 

62.50 

.8892 

20 

16.28 

.9751 

71 

63.58 

.8867 

21 

17.11 

.9741 

72 

64.66 

.8842 

22 

17.95 

.9731 

73 

65.74 

.8817 

23 

18.78             .9720 

74 

66.83 

.8791 

24 

19.62             .9710 

75 

67.93 

.8765 

25 

20.46        i     .9700 

76 

69.05 

.8739 

26 

21.30        i     .9689 

77 

70.18 

.8712 

27 

22.14              9679 

78 

71.31 

.8685 

28 

22.99             .9668 

79 

72.45 

.8658 

29 

23.84 

.9657 

80 

7359 

.8631 

•  30 

24.69 

.9646 

81 

74.74 

.8603 

31 

25.55 

.9634 

82 

75.91 

.8575 

32 

26.41 

.9622 

83 

77.09 

.8547 

33 

27  27 

.9609 

84 

78.29 

.8518 

34 

28.13 

.9596 

85 

79.50 

.8488 

35 

28.99 

.9583 

86 

80.71 

.8458 

36 

29.86 

.9570 

87 

81.94 

.8428 

37 

30.74 

.9556 

88 

83.19 

.8397 

38 

31.62 

.9541 

89 

84.46 

.8365 

39 

32.50 

.9526 

90 

85.75 

.8332 

40 

33.39 

.9510 

91 

87.09 

.8299 

41 

34.28 

.9494 

92 

83.37 

.8265 

42 

35.18 

.9478 

93 

89.71 

.8230 

43 

36.08 

.9461 

94 

91.07 

.8194 

44 

36.99 

.9444 

95 

92.46 

.8157 

45 

37.90 

.9427 

96 

93.89 

.8118 

46 

38.82 

.9409 

97 

95.34 

.8077 

47 

39.75 

.9391 

98 

96.84 

.8034 

48 

40.66 

.9373 

99 

98.39 

.7988 

49 

41.59 

.9354 

100 

100.00 

.7939 

50 

42.52 

.9335 

Tralles. 


SPECIFIC   GRAVITY   OF   SULPHURIC   ACID. 


171 


PERCENTAGE    AND    SPECIFIC   GRAVITY   OF    SULPHURIC 
ACID  AT   15°  C*     WATER   AT    0°  C.  =  1. 


Percent. 

Sp.  Gr. 

Percent. 

Sp.  Gr. 

Percent. 

Sp.  Gr. 

Percent. 

Sp.  Gr. 

1 

l.OOG 

26 

1.191 

51 

1.409 

76 

1.684 

2 

1.012 

27 

1.199 

52 

1.418 

77 

1.697 

3 

1.018 

28 

1.207 

53 

1.428 

78 

1.710 

4 

1.025 

29 

1.215 

54 

1.438 

79 

1.721 

5 

1.032 

30 

1.223 

55 

1.448 

80 

1.732 

6 

1.039 

31 

1.231 

56 

1.459 

81 

1.743 

7 

1.046 

32 

1  239 

57 

1.469 

82 

1.753 

8 

1.053 

33 

1.247 

58 

1.480 

83 

1.763 

9 

1.061 

34 

1.256 

59 

1.491 

84 

1.773 

10 

1.069 

35 

1.264 

60 

1.501 

85 

1.783 

11 

1.076 

36 

1.272 

61 

1.512 

86 

1.792 

12 

1.084 

37 

1.281 

62 

1.523 

87 

1.800 

13 

1.091 

38 

1.290 

63 

1.535 

88 

1.807 

14 

1.099 

39 

1.298 

64 

1.546 

89 

1.814 

15 

1.106 

40 

1.307 

65 

1.558 

90 

1.820 

16 

1.114 

41 

1.316 

66 

1.569 

91 

1.825 

17 

1.122 

42 

1.324 

67 

1.580 

92 

1.8294 

18 

1.129 

43 

1.333 

68 

1.592 

93 

1.8339 

19 

1.137 

44 

1.342L 

69 

1.604 

94 

1.8372 

20 

1.145 

45 

1.352 

70 

1.615 

95 

1.8390 

21 

1.153 

46 

1.361 

71 

1.626 

96 

1.8406 

22 

1.161 

47 

1.370 

72 

1.638 

97 

1.8410 

23 

1.168 

48 

1.379 

73 

1.650 

98 

1.8412 

24 

1.176 

49 

1.389 

74 

1.662 

99 

1.8403 

25 

1.184 

50 

1.399 

75 

1.674 

100 

1.8384 

*  GerlachZ.  Anal.  Chem.  27,316. 


172 


SPECIFIC   GRAVITY   OF    NITRIC   ACID. 


PERCENTAGE  AND   SPECIFIC  GRAVITY   OF   NITRIC   ACID 
AT  15°C  * 


Per  Cent. 

Sp.  Gr. 

Per  Cent. 

Sp.  Gr. 

Per  Cent. 

Sp.  Gr. 

Per  Cent. 

Sp.  Gr. 

1 

1.006 

26 

1.159 

51 

1.323 

76 

1.445 

2 

1.012 

27 

1.166 

52 

1.329 

77 

1.449 

3 

1.018 

28 

1.172 

53 

1.335 

78 

1.452 

4 

1.024 

29 

1.179 

54 

1.341 

.79 

1.456 

5 

1.029 

30 

1.185 

55 

1.346 

80 

1.460 

6 

1.035 

31 

1.192 

56 

1.352 

81 

1.463 

7 

1.040 

32 

1.198 

57 

1.358 

82 

1.467 

8 

1.045 

33 

1.204 

58 

1.363 

83 

1.470 

9 

1.051 

34 

1.210 

59 

1.369 

84 

1.474 

10 

1.057 

35 

1.218 

60 

1.374 

85 

1.478 

11 

1.064 

36 

1.225 

61 

1.380 

86 

1.481 

12 

1.070 

37 

1.230 

62 

1.386 

87 

1.484 

13 

1.077 

38 

1.236 

63 

1.390 

88 

1.488 

14 

1.083 

39 

1.244 

64 

1.395 

89 

1.491 

15 

1.089 

40 

1.251 

•  65 

1.400 

90 

1.495 

16 

1.095 

41 

1.257 

66 

1.405 

91 

1.499 

17 

1.100 

42 

1.264 

67 

1.410 

92  ' 

1.503 

18 

1.106 

43 

1.270 

68 

1.414 

93 

1.506 

19 

1.112 

44 

1.276 

69 

1.419 

94 

1.509 

20 

1.120 

45 

1.284 

70 

1.423 

95 

1.512 

21 

1.126 

46 

1.290 

71 

1.427 

96 

1.516 

22 

1.132 

47 

1.298 

72 

1.431 

97 

1.520 

23 

1.138 

48 

1.304 

73 

1.435 

98 

1.523 

24 

1.145 

49 

1.312 

74 

1.439 

99 

1.526 

25 

1.151 

50 

1.316 

75 

1.442 

100 

1.530 

*  Kolb,  Gerlach,  Z.  Anal.  Chem.  8,292. 


SPECIFIC   GRAVITY   OF    HYDROCHLORIC   ACID. 


173 


SPECIFIC   GRAVITY   AND    PERCENTAGE   OF   HYDRO- 
CHLORIC ACID  AT  15°  C*     WATER  AT  4°  C  =  1. 


Sp.  Gr. 

%HC1. 

Sp.  Gr. 

%  HC1. 

1.000 

0.16 

1.105 

20.97 

1.005 

1.15 

1.110 

21.92 

1.010 

2.14 

1.115 

22.86 

1.015 

3.12 

1.120 

23.82 

1.020 

4.13 

1.125 

24.78 

1.025 

5.15 

1.130 

25.75 

1.030 

6.15 

1.135 

26.70 

1.035 

7.15 

1.140 

27.66 

1.040 

8.16 

1.145 

28.61 

1.045 

9.16 

1.150 

29.57 

1.050 

10.17 

1.155 

30.55 

1.055 

11.18 

1.160 

31.52 

1.060 

12.19 

1.165 

32.49 

1.065 

13.19 

1.170 

33.46 

1.070 

14.17 

1.175 

34.42 

1.075 

15.16 

1.180 

35.39 

1.080 

16.15 

1.185 

.       36.31 

1.085 

17.13 

1.190 

37.23 

1.090 

18.11 

1.195 

38.16 

1.095 

19.06 

1.200 

39.11 

1.100 

20.01 

For  other  similar  tables  see  Comey's  Dictionary  of  Chemical 
'Solubilities. 


*  Lunge  and  Marchlewski  Zeit.  f.  angew.  Chem.  1891.  133. 


174 


SPECIFIC   GRAVITY   OF   AMMONIA. 


SPECIFIC  GRAVITY  AND   PERCENTAGE   OF 
AMMONIA*   AT   15°    C.     • 


Sp.  Gr. 

%NHS 

Sp.  Gr. 

%NH, 

Sp.  Gr. 

KNH, 

1.000 

0.00 

0.960 

9.91 

0.920 

21.75 

0.998 

0.45 

0.958 

10.47 

0.918 

22.39 

0.996 

0.91 

0.956 

11.03 

0.916 

23.03 

0.994 

1.37 

0.954 

11.60 

0.914 

23.68 

0.992 

1.84 

0.952 

12.17 

0.912 

24.33 

0.990 

2.31 

0.950 

12.74 

0.910 

24.99 

0.988 

2.80 

0.948 

13.31 

0.908 

25.65 

0.986 

3  30 

0.946 

13.88 

0.906 

26.31 

0.984 

3.80 

0.944 

14.46 

0.904 

26.98 

0.982 

4.30 

0.942 

15.04 

0.902 

27.65 

0.980          4.80 

0.940 

15.63 

0.900 

28.33 

0.978 

5.30 

0.938 

16.22 

0898 

29.01 

0.976 

5.80 

0.936 

16.82 

0.896 

29.69 

0.974 

6.30 

0.934 

17.42 

0.894 

30.37 

0.972 

6.80 

0.932 

18.03 

0.892 

31.05 

0.970 

7.31 

0.930 

18.64 

0.890 

31.75 

0.968 

7.82 

0.928 

19.25 

0.888 

32.50 

0.966           8.33 

0.926 

19.87 

0.886 

33.25 

0.964          8.84 

0.924 

20.49 

0.884 

34.10 

0.962           935 

0.922 

21.12 

0.882 

34.95 

*  Lunge  and  Wiernik  Zeit.  f.  angew.  Chem.  1889.  183. 


ELECTRICAL  UNITS. 


175 


TABLE  SHOWING   THE   RELATIONS   BETWEEN    UNITS   OF 
ELECTRICITY,  HEAT  AND  POWER. 


1  ampere  =  1  volt  -i-  1  ohm. 

=  1  coulomb  per  second. 
1  ampere  hour  =  1  coulomb  per  second 
kept  up  for  one  hour. 
1  ampere  hour  =3600  coulombs. 
1  volt  =  1  ampere  X  1  ohm. 
1  ohm  =  1  volt  -r-  1  ampere. 
1  joule  =  1  volt  X  1  coulomb. 
"      =  .2381  calorie. 

=  .73732  foot-pound. 
"      =  .10194  kilogram-meter. 
1  calorie  =4.2  joules. 

—  3.0968  foot-pounds. 
=  .42815  kilogram-meter. 
1  foot  pound  =  1.3563  joules. 
=  .32292  calorie. 
=  .13825  kilogram-meter 
1  kilogram  meter  =  9.81  joules. 

=  2.3362  calories. 
=  7.233  foot-pounds. 
1  watt-hour  =  1  watt  kept  up  for  one 

hour. 
1  watt-hour  =  1  joule  per  second  kept 

up  for  one  hour. 
1  watt-hour  =  3600  joules. 

=  857.16  calories. 
=  2654.4  foot-pounds. 
=  366. 98  kilogram-meters 
1  watt  =  1  joule  per  second. 

"     =  .2381  calorie  per  second. 

"     =  .  73732  foot-pound  per  second . 


1  watt  =.10194   kilogram-meter    per 

second. 
1  watt  =  .0013406  horse-power. 

"     =  .001  kilowatt. 
1  horse-power  =  745.94  watts. 

=  177. 6  calories  per  sec- 
ond. 
1  horse-power  =    10656.    calories    per 

minute. 
1  horse -power  =  550   foot  pounds  per 

second. 
1  horse-power  =  33000  foot  pounds  per 

minute. 

1  horse-power  =  76.04  kilogram-met- 
ers per  second. 

1  horse-power  =  4562.4  kilogram-met- 
ers per  minute. 

1  horse-power  =  .74594  kilowatts. 
1  kilowatt  =  1000  watts. 

"         =  1000  joules  per  second. 
"         =  238.1  calories  per  second. 
' '         =  14286.  calories  per  minute 
"         =  737.32   foot-pounds   per 

second. 
1  kilowatt  =  44239.    foot-pounds  per 

minute. 
1  kilowatt  =  101.94    kilogram-meters 

per  second. 
1  kilowatt  =  6116.4    kilogram-meters 

per  minute. 
1  kilowatt  =  1.3406  horse-power. 


176          TABLE    OF    ELECTRO-CHEMICAL    EQUIVALENTS. 


Table  of  Electro-Chemical  Equivalents  based  on  the  definition  of  the 
ampere  and  the  atomic  weights,  oxygen  being  16,  as  given 
on  page  1 59. 


1 

* 

1 

3~-  ? 

jjj.1 

11 

M 

ll 

Element. 

"E 
o 

? 

1" 

ja 
U 

£w  2 
3    o 

"5  ? 
dl 

I  8. 

O  E 
rt 

£!  £. 
f* 

Electropositive. 

27  11 

^ 

9  037 

00009362 

10680. 

3370 

1346 

120  43 

^ 

40  14 

0004159 

2405 

1  497 

303  0 

Bismuth  

208  11 

3 

69.37 

.0007186 

1392. 

2587 

175.3 

112  38 

9: 

56  19 

0005821 

1718 

2.096 

216  4 

Cobalt  
Copper  (Cuprous)  
Copper  (Cupric)  
Gold  
Hydrogen  
Iron  (Ferrous)  
Iron  (Ferrie)  
Lead 

58.99 
63.6 
63.6 
197.23 
1.008 
56.02 
56.02 
90g  92 

2. 
1. 
2. 
3. 
1. 
2. 
3. 
2 

29.5 
63.6 

31.8 
65.78 
1.008 
28.01 
18.67 
103  5 

.0003056 
.000658VJ 
.0003295 
.0006815 
.00001044 
.0002902 
.0001934 
001072 

3272. 
1518. 
3036. 
1467. 
95785. 
3446. 
5171. 
932  8 

1.100 
2.372 
1.186 
2.453 
.03758 
1.045 
.6962 
3  859 

412.2 
191.2 
382.4 
184.7 
12070. 
434.0 
651.5 
117  5 

Magnesium  

24.28 
54  99 

2. 

0 

12.14 
27  50 

.0001258 
0002849 

7949. 
3510 

.4528 
1  026 

1001. 
442  1 

Mercury  (Mercurous).  .  .  . 
Mercury  (Mercuric)  
Nickel  
Platinum  . 

200. 
200. 
58.69 
194  89 

1. 
2. 
2, 
4 

200. 
100. 
29.35 

48  72 

.002072 
.001036 
.0003041 
0005047 

482.6 
965.3 
3288. 
1981 

7.459 
3.730 
1.095 
1.817 

60.81 
121.6 
414.2 
2496 

Potassium. 

39  11 

1 

3911 

0004052 

2468 

1.459 

3109 

Silver  
Sodium  

10792 
2305 

1. 
1 

107.92 
23.05 

.001118 
0002388 

894.5 
4188. 

4.025 
.8597 

112.7 
527.6 

Tin  (Stannous)  
Tin  (Stannic;  
Zinc  . 

119.05 
119.05 
65  41 

2. 

4. 

9, 

59.52 
29.76 
3270 

.0006166 
.0003083 
0003388 

1622. 
3243. 
2952 

2.220 
1.110 
1.220 

204.3 
408.6 
371.8 

Electronegative. 
Bromine 

79  95 

1 

79  95 

0008283 

1207. 

2.982 

152.1 

Chlorine  
Iodine  
N  itrogen  

35.45 
126.85 
14.04 

1. 
1. 

3 

35.45 

126.85 
4.68 

.0003673 
.001314 
.00004848 

2723. 
761.0 

20627. 

1.322 
4.730 
.1745 

343.1 
95.90 
2599. 

Oxygen. 

16. 

<>, 

8 

.00008288 

12066. 

.2984 

1520. 

HEATS   OF   COMBINATION. 


177 


Heats  of    Combination    in   Calories,     for   Equivalent    Weights  in 
grams,  of  Chlorides,  Bromides,  Iodides,  Sulphates  and  Nitrates.* 


Element. 

Valence. 

Chloride. 

Bromide. 

Iodide. 

Sulphate 

Nitrate. 

Aluminium 

3 

53660 

39900 

23463 

25315 

Antimony 

3      ;    30463 

Bismuth 

3          30210 

Cadmium 

2         46620 

37600 

24215 

44940 

43000 

Cobalt 

2 

38240 

44350 

42270 

Copper 

1 

32875 

24985 

16260 

Copper 

2 

27980 

26205 

Gold 

3           7607 

2950 

Hydrogen 

1      !    39315 

28380 

13170 

39170 

33830 

Iron 

2         41025 

46600 

44835 

Lead 

2 

41385 

32225 

19900 

34Q35 

Magnesium 

2 

75505 

90090 

88240 

Manganese 

2 

55995 

60625 

58860 

Mercury 

2 

31580 

17155 

18535 

Nickel 

2 

37265 

43475 

41710 

Silver 

1 

29380 

22700 

13800 

10195 

8390 

Tin 

2 

40395 

Zinc 

2 

48605 

37965 

24615 

53045 

51255 

*  These  values  are  taken  from  Thomson's  Thermo-chemische 
Untersuchungen.  The  sulphates,  nitrates  and  hydrogen  com- 
pounds are  for  aqueous  solutions.  The  chlorides,  bromides  and 
iodides  are  anhydrous. 


178  SPECIFIC    RESISTANCE. 

SPECIFIC  RESISTANCE  OF  VARIOUS  SUBSTANCES. 


Substance. 

Temperature. 

>> 

> 

o  2 

£0 

1 

p 
in  ohms. 

§ 

1 
^ 

Observer. 

Silver  (annealed)  

0°C. 

10  6 

.00000150< 

) 

Matthiesen. 

Silver  (hard  drawn)  
Copper  (annealed)  

(l 

10.6 
8  9 

.000001531 
.  00000159' 

) 

I 

Copper  (hard  drawn)  

" 

8  9 

.00000162" 

) 

,i 

Gold  (annealed)  
Gold  (hard  drawn)  

** 

19.3 
19  3 

.00000205S 
00000208' 

J 

l 

It 

Aluminium  (annealed)  
Platinum  (annealed)  

'• 

2.6 

21.2 
8  1 

.00000290J 
.00000903( 
00000968' 

J 
) 

r 

:: 

•  i 

7  3 

00001317 

IS 

7  1 

00000559* 

$ 

,4 

it 

11  4 

00001957 

,, 

i. 

8  5 

00001242 

„ 

n 

about   00002 

09 

,, 

Graphite           

2  3 

.0024  to  .fr 

13 

Everett 

Retort  Carbon       

•i 

1  9 

.07 

Mercury 

13  6 

.00009407. 

!* 

Nitric  Acid  in  water  
Hydrochloric  Acid  in  water.  .  .  . 

18°C. 

1.185 
1.092 
1  224 

1.28 
1.31 
1  36 

29.7 
18.3 
SO  4 

Kohlrausch. 

Phosphoric  Acid  in  water  
Tartaric  Acid  in  water  
Acetic  Acid  in  water  

'•« 

1.307 
1.107 
1.022 

4.79 
9.97 
61.9 

46.8 
22.4 
16  6 

'•' 

Ammonium  Chloride  in  water.  . 
Sodium  Chloride  in  water  
Sodium  Sulphate  in  water  

10°C. 

1.270 

2.5 

4.7 
11.3 
28.5 

1       Kohlrausch 
and 
Nippoldt. 
Ewing  &  Macgregor 

II 

1.422 

33  7 

Copper  Sulphate  in  water  
Potassium  Sulphate  in  water.  .  . 
Potassium  Bichromate  in  water. 

.. 

1.205 

29.3 
16.6 
29.6 

: 

.. 

Further  data  on  the  conductivity  of  solutions  can  be  found  in  Physikalisch-Chem- 
ische  Tabellen,  Landolt  and  Bernstein,  pp.  103, 106. 

Kohlrausch  and  Nippoldt,  Fogg.  Ann.  138  p.  379,  1869. 

Grotrian  "          "      151  p.  378,  1874. 

Kohlrausch  and  Grotrian,  Pogg.  Ann.  159  p.  238,  1876  and  Wied.  Ann.  6,  p.  145, 
1879. 


*  From  the  definition  of  the  ohm. 
f  Solution  of  minimum  resistance, 
f  Saturated  solution. 


COMPARISON   OF   VOLTAGES. 


17!'' 


Comparison  of  Calculated  and  Observed  Decomposition  Voltages  of 
various  Aqueous  Solutions. 


Chloride. 

Bromide. 

Iodide. 

Sulphate. 

Nitrate. 

j 

1 

a 

o 

Calculated. 

Observed. 

1 

o 

Calculated. 

j 

Magnesium. 

3.28 

3.1 

2.56 

2.01 

3.91 

3.83 

Zinc.    .  .   . 

2.11 

2.11 

1.65 

1.79 

1.07 

1.25 

2.31 

2.35 

2.23 

Cadmium.  . 

2.03 

1.9 

1.63 

1.58 

1.06 

1.12 

1.95 

2.03 

1.87 

1.98- 

Aluminium. 

2.33 

2.0 

1.74 

1.53 

1.02 

.88 

1.10 

Iron.    .   .    . 

1.78 

1.6 

1.30 

.68 

2.03 

1.95 

Cobalt.    .   . 

1.66 

1.43 

1.05 

.51 

1.91 

1.92 

1.84 

Nickel.  .   .   . 

1.62 

1.33 

.85 

.36 

1.89 

2.09 

1.81 

Tin  

1.76 

1.61 

1.30 

.71 

Lead.  .  .   . 

1.80 

1.63 

1.40 

1.33 

.87 

.83 

1.48 

1.52 

Copper.    .    . 

1.43 

1.32 

1.09 

1  02 

.71 

.64 

1.22 

1.14 

Silver.    .    . 

1.28 

1.11 

.99 

.95 

.60 

.65 

.44 

.365 

.36* 

Antimony.  . 

1.32 

1.22 

.80 

.44 

Bismuth.   . 

1.31 

1.21 

.92 

.43 

Hydrogen.  . 

1.71 

1.31 

1.23 

.94 

.57 

.52 

1.70 

1.67 

1.81 

1.69 

The  observed  decomposition  values  of  the  acids  given 
under  hydrogen  together  with  many  of  the  sulphates  and 
nitrates  are  taken  from  LeBlanc's  Electro  Chemistry  pp.  247-8. 
The  other  observed  values  are  from  Crocker,  Trans.  Am.  Inst. 
E.E.  1885,  p.  281. 

The  calculated  values   are   obtained  by   substituting-  the- 

TT 

heats  of  combination  on  page  177  in  equation  (14)   E  = 

-coUUU. 


*  Experiment. 


180           LOGARITHMS  OF  NUMBERS. 

|J 

o 

1 

O 

Q 

4" 

7 

Q 

Proportional  Parts. 

2  £ 

ij 

\J 

J. 

& 

O 

I 

o 

1  1  1  2  |  I 

4 

5|6 

7 

8 

If 

10 

0000 

0043 

0036 

0128 

0170 

0212 

0253 

0294 

0334 

03741  4:  8 

]L 

17 

I 

21|25 

29 

33 

fc 

11 

0414 

0453 

0492 

0531 

0569 

OG07 

0645 

0682 

0719 

0755  §  4J  8 

11 

13 

19  23 

•2C, 

30  34 

12 

0792 

0328 

0864 

0399 

0934 

0969 

1004 

1038 

1072 

1106 

3 

7 

10 

11 

17 

21 

24 

28 

3! 

13 

11159 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

3 

fi 

10 

13 

10 

13 

23 

26 

29 

14 

146] 

1492 

.1523 

1553 

1584 

1614 

1644 

1673  1703 

1732 

3  6 

9|12 

15 

18 

21 

2i 

27 

15 

1761 

17SO 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

3!   Q 

811 

14 

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This  book  is  DUE  on  the  last  rt  date  stamped 

date  stamped  below 


I  OCT   8      194- 


NOV  29 


AU62      1948 
2 


APR  2  3  1952. 
AUG  13 


1, 


5m-2,'31 


. 


Sic         Calculations 


chemistr. 


V\(o\C 


UNIVERSITY  of  CALIFORNIA 

LOS  ANGELES 
LIBRAEY 


